Principles of Calculus

In the year one, we learnt about the concept of limits and differentiating functions by first principles and by using the power rule. In this section, we will learn about the power, product, quotient and chain rules to find derivatives of functions, which are crucial for solving real-world problems involving rates of change. We will also cover Implicit differentiation which allows us to handle equations with interrelated variables and differentiating transcendental functions, such as exponential, logarithmic and trigonometric functions, which broadens your mathematical toolkit. These topics collectively build a solid foundation for advanced studies and diverse career applications.

KEY IDEAS

• Chain Rule: If f(g(x)), let u = g(x), so y = f(u), then dy___ dx = dy___ du × du__ dx

• Derivative of eˣis eˣand if g(x) = eᶠ(x) then d__ dx(g(x)) = f′(x) × eᶠ(x)

• Derivative of cosx is − sinx

• Derivative of sinx is cosx

• Derivative of the natural logarithm, ln(x), is 1/x and if f(x) = logₐ(g(x)) then d__ dx fx = 1___ lna × 1__ gx × g′(x)

• Derivative represents the rate of change of a function with respect to its input.

• Differentiability is a function which is differentiable if its derivative exists at a point.

• Limit: the derivative of a function f(x) is defined as the limit of the difference quotient.

• Power Rule: If y = axⁿ, where a is a constant, then dy__ dx = na xⁿ⁻¹• Product Rule: If y = uv , dy__ dx = v du___ dx + u dv___ dx

• Quotient Rule: if y = u/v where u and v are function of x , then dy___ dx = v du___ dx − u dv__ dx________

Differentiation is a process to find the derivative or gradient function. There are a number of rules associated with differentiation. These rules can be used to differentiate more complicated functions without having to use the first principles.

The derivative of y with respect to x is usually written as dy___ dx . This is called the Leibniz notation. The derivative of f(x) with respect to x is usually written as f′(x) or dy___ dx [f(x)].

Figure 10.1: Graphical illustration of ∆ y___ ∆ x The gradient (m) of the sloping straight line shown in the diagram above is defined as:

m = the vertical distance the line rises or falls between two points P and Q_________________________________________________ the horizontal distance between P and Q Where P is the point to the left of point Q on the straight-line AB which slopes upwards from left to right. The changes in the x and y values of the point P and Q are denoted by dx and dy respectively. The gradient of the line is dy___ dx .

We could have chosen any pair of points on the straight line for P and Q and by similar triangles; this ratio would have been worked out to the same value. The gradient of a straight line is constant throughout its’ length. Its value is denoted by the symbol m. Therefore m = dy___ dx .

For example, P is the point (2, 3) and Q is the point (6, 4), then P is to the left and below the point Q.

dy = the change in the y− values = 4 − 3 = 1 dx = the change in the x− values = 6 − 2 = 4 m = dy___ dx = 1/4 = 0.25

Note: the sloping line rises vertically from left to right by 0.25 unit for every 1 unit horizontally.

Figure 10.2: Graphical illustration of gradient of line PQ If for another line M is the point ( − 1, 3) and N is the point (0, − 4), then M is to the left and above the point M.

dy = the change in the y− values = 3 − − 4 = 7 dx = the change in the x− values = − 1 − 0 = − 1 m = dy___ dx = 7__

–1 = − 7

Figure 10.3: Graphical illustration of gradient of line MN The lines going up to the right have a positive gradient, lines going down to the right have a negative gradient.

Rules of differentiation Certain rules have been made to simplify and guide the process of differentiation.

Let us work through these rules together.

Differentiation of a constant function If y = c where c is constant, then dy__ dx = 0 . Remember if y = c then this is a horizontal line, and, like all horizontal lines, the gradient is zero.

For example, if y = 5, dy___ dx = 0 Therefore, the derivative of any constant is zero.

Differentiation of ax where a is a constant If y = ax where a is a constant, then dy__ dx = a .

For example, if y = 5x , then dy__ dx = 5 Also, if y = − 2x, then dy__ dx = − 2 Differentiation of xⁿIf y = xⁿ, then dy__ dx = n xⁿ⁻¹, where n is any real number. Therefore, the general case for differentiating xⁿis to multiple the term by the power (exponent) and decrease the power by one.

For example, if y = x⁴, then dy__ dx = 4 x³.

Also, if y = x⁶, then dy__ dx = 6 x⁵.

Differentiation of axⁿIf y = axⁿ, where a is a constant, then dy__ dx = na xⁿ⁻¹.

For example, if y = 6x³, then dy__ dx = 18x².

Also, if y = 4x⁻³, then dy__ dx = − 12x⁻⁴.

This is known as the power rule.

Derivative of the sum and difference The differential coefficient of sum and difference is the sum and difference of the differential coefficients of the separate terms. Thus differentiate the sum and difference term by term.

Hence, d__ dx[ f(x) + g(x) − h(x)] = d__ dx[ f(x)] + d__ dx[g (x)] − d______ dx[h(x)].

Let us go through some examples to practise this rule.

Example 10.1

Find the derivative function with respect to x .

a. 5 x²+ 3x

b. 2 x³− 6x²+ 4x − 2

c. 6 x ⁻¹___ 3 − 2x

d. 3 x²+ 7 − 4/x

Solution

a. dy__ dx = d__ dx( 5x²) + d__ dx(3x) = 10x + 3

b. dy__ dx = d__ dx(2x³) − d__ dx(6x²) + d__ dx(4x) − d__ dx(2) = 6x²− 12x + 4

c. dy__ dx = d__ dx( 6x ⁻¹__ 3 ) − d__ dx(2x) = 6(−1/3) x ⁻⁴___ 3 − 2 = − 2 x ⁻⁴___ 3 − 2

d. dy__ dx = d__ dx(3x²)) + d__ dx(7 ) − d__ dx (4x⁻¹) = 6x + 0 − ( − 4) x⁻²= 6x + 4/x²Example 10.2 Find dy__ dx if:

a. y = 4/x³b. y = 1___ √_ x

c. y = 1/x⁵d. y = 3 √x

Solution

a. y = 4/x³= 4x⁻³dy__ dx = − 12x⁻⁴b. y = 1___ √_ x = 1/x¹__ 2 = x −1/2 dy__ dx = −1/2 x −3/2 = − 1/2 √x³c. y = 1/x⁵= x⁻⁵dy_ dx = − 5 x⁻⁶= − 5/x⁶d. y = 3√x = 3x ¹__ 2 dy__ dx = 3/2 x −1/2 = 3/2 √x

Note: If the expression is such that it cannot be easily multiplied out or divided into separate terms the product or quotient rules must be used.

Derivative of linear combination of functions The derivative of a linear combination of functions can be found using the linearity property of differentiation. This property states that the derivative of a linear combination of functions is equal to the linear combination of the derivatives of the derivatives of those functions Definition:

If you have two differentiable functions f(x) and g(x) and constants a and b the linear combination of these functions is given by:

h(x) = af(x) + bg(x) Derivative:

The derivative of h(x) is:

hᴵ(x) = d__ dx[af(x) + bg(x)] Using the linearity property, we can separate the terms:

hᴵˣ= a d__ dx[f(x)] + b d__ dx[g(x)] Simplify to get the result:

a × f′(x) + b × g′(x)

Example 10.3

Find the derivative of f(x) = 4x³+ 7x²Solution h(x) = 4x³+ 7x²f(x) = x³and g(x) = x²Constants a = 4 and b = 7 Derivative of Individual functions:

f ᴵ(x) = d__ dx (x³) = 3x²gᴵ(x) = d__ dx (x²) = 2x Applying the linearity property:

hᴵ(x) = a fᴵ(x) + b gᴵ(x) hᴵ(x) = 4 × 3x²+ 7 × 2x hᴵ(x) = 12 x²+ 14x

Example 10.4

Find the derivative of f(x) = 5x⁴− 3x³+ 2x − 1

Solution

f(x) = 5x⁴− 3x³+ 2x − 1 u(x) = x⁴, v(x) = x³, w(x) = x and c = − 1 Constants: a = 5 , b = − 3, d = 2 Derivative of Individual functions:

uᴵ(x) = d__ dx (x ⁴) = 4x ³vᴵ(x) = d__ dx (x ³) = 3x ²wᴵ(x) = d__ dx(x) = 1 The derivative of the constant is 0 Applying the linearity property:

fᴵ(x) = a uᴵ(x) + b vᴵ(x) + d wᴵ(x) fᴵ(x) = 5 × 4x³− 3 × 3x²+ 2 × 1 fᴵ(x) = 20x³− 9 x²+ 2

Example 10.5

Find the derivative of f(x) = 7x⁴− 5x³+ 3x − 4

Solution

f(x) = 7x⁴− 5x³+ 3x − 4 u(x) = x⁴, v(x) = x³, w(x) = x and c = − 4 (constant term) Constants: a = 7 , b = − 5, d = 3 Derivative of Individual functions:

uᴵ(x) = d__ dx (x ⁴) = 4x ³vᴵ(x) = d__ dx (x ³) = 3x ²wᴵ(x) = d__ dx(x) = 1 The derivative of the constant is 0 Applying the linearity property:

fᴵ(x) = a uᴵ(x) + b vᴵ(x) + d wᴵ(x) fᴵ(x) = 7 × 4x³− 5 × 3x²+ 3 × 1 fᴵ(x) = 28x³− 15 x²+ 3 Chain Rule (Composite function) A function such as y = 1_______ (3x − 4)²= (3x − 4)⁻²is a problem for differentiation. It cannot be expressed as separate terms in a polynomial. To differentiate this we need to use the chain rule. To find dy__ dx, we treat (3x − 4)⁻²as composite function, built up in two stages from a core function (3x − 4) which we will call u and then taking y = u⁻². We have u = 3x − 4 and y = u⁻².

If y = g(u) where u = f(x), then dy__ dx = dy___ du × du__ dx This rule is extremely important and enables us to differentiate complicated functions.

Note: du cannot be ‘cancelled’ on the right-hand side as these are not fractions but derivatives.

Example 10.6

Find dy__ dx if y = (2x − 1)³Solution y = (2x − 1)³Let u = 2x − 1 and y = u³du__ dx = 2 and dy__ du = 3u² But dy__ dx = y__ du × du__ dx = 3 u²× 2 But u = 2x − 1 , so we substitute this in:

dy__ dx = 3 (2x − 1)²× 2 = 6( 2x − 1)²Example 10.7 Find dy__ dx if y = ( x²+ 3)⁻¹Solution y = ( x²+ 3)⁻¹Let u = x²+ 3 and y = u⁻¹du__ dx = 2x and dy__ du = − u⁻²But dy__ dx = y__ du × du__ dx = − u⁻²× 2x, but u = x²+ 3 dy__ dx = −(x²+ 3)⁻²× 2x = − 2x (x²+ 3)⁻²= − x_______ ( x²+ 3)²Example 10.8 Find dy__ dx if y = √_____ 2x + 5

Solution

y = √2x + 5 = (2x + 5)¹__ 2 Let u = 2x + 5 and y = u ¹__ 2 du__ dx = 2 and dy__ du = 1/2 u −1/2 But dy__ dx = dy__ du × du__ dx = 1/2 u⁻¹___ 2 × 2 = 1/2( 2x + 5)⁻¹___ 2 × 2 = ( 2x + 5)⁻¹___ 2 = 1_______ √_____ 2x + 5 Product rule y = (3x − 1)³( x²+ 5) is the product of two expressions, (3x − 1)³and ( x²+ 5). To find dy__ dx, we could expand the product and then differentiate each term separately, but this would be time consuming. Therefore, let us see how dy__ dx can be found without doing this.

Let y = uv where u and v are functions of x . Then uv is the product of two functions.

In the above, for example, u would be (3x − 1) and v would be ( x²+ 5).

Take an increment δx in x which will in turn produce increments δu in u and δv in v , finally producing a change δy in y . Then:

y + δy = (u + δu)(v + δv) = uv + vδu + uδv + (δu)(δv), so δy = vδu + uδv + (δu)(δv), and δy__ δx = v δu__ δx + u δv__ δx + δu__ δx δv Now, let δx → 0 , then δu → 0 as a result.

∴ δu__ δx → du__ dx, δv__ δx → dv__ dx and δy__ δx → dy__ dx So, the limiting value dy__ dx = v du__ dx + u dv__ dx, (the third term → 0 as δu__ δx is definite, but δv → 0).

Hence for y = uv , we have the product rule: dy__ dx = v du__ dx + u dv__ dx

Example 10.9

Find the derivate of y = (3x − 2)( x²+ 3)

Solution

Let u = (3x − 2) and v = ( x²+ 3) du__ dx = 3 and dv__ dx = 2x dy_ dx = v du___ dx + u dv___ dx dy__ dx = (x²+ 3)(3) + (3x − 2)(2x) = 3x²+ 9 + 6x²− 4x = 9x²− 4x + 9

Example 10.10

Differentiate (x²)( x + 1)⁵with respect to x

Solution

y = (x²)( x + 1)⁵Let u = ( x²) and v = ( x + 1)⁵du__ dx = 2x and dv__ dx = 5( x + 1)⁴, differentiated using the chain rule dy__ dx = v du__ dx + u dv__ dx = x²[5(x + 1)⁴] + (x + 1)⁵(2x) dy_ dx = 5x²( x + 1)⁴+ ( x + 1)⁵(2x) = 5x²( x + 1)⁴+ 2x( x + 1)⁵We could then factorise this and simplify:

dy_ dx = x (x + 1)⁴[5x + 2(x + 1)] = x (x + 1)⁴(7x + 2)

Example 10.11

Differentiate (x + 1)³(2 x − 5)²with respect to x

Solution

y = ( x + 1)³(2 x − 5)²Let u = ( x + 1)³and v = (2 x − 5)²du__ dx = 3( x + 1)²and dv__ dx = 2(2x − 5)(2), differentiated using the chain rule dy__ dx = v du__ dx + u dv__ dx = (2x − 5)²(3) (x + 1)²+ ( x + 1)³(4)(2x − 5) dy__ dx = 3( 2x − 5)²( x + 1)²+ 4(x + 1)³(2x − 5) We could then factorise this and simplify:

dy_ dx = (2x − 5) (x + 1)²[3(2x − 5) + 4(x + 1)] = (2x − 5) (x + 1)²(10x − 11) Quotient Rule Expression like x²+ 1_____ 2x − 5, √_ x/1 − 5x, and x²______ (x − 4)³are called quotients because they represent the division of one function by another.

Quotient functions have the form Q(x) = u(x)___ v(x) Notice that u(x) = Q(x)v(x) ∴ u′(x) = Q′(x)v(x) + Q(x)v′(x), using the Product Rule ∴ u′(x)− Q(x)v′(x) = Q′(x)v(x) Q′(x)v(x) = u′(x)− u(x)____ v(x) v′(x) Q′(x)v(x) = u′(x)v(x) − u(x)v′(x)______________ v(x) Q′(x) = u′(x)v(x) − u(x)v′(x)______________ [v(x)]², when this exists If Q(x) = u(x)___ v(x) then Q′(x) = u′(x)v(x) − u(x)v′(x)______________ [v(x)]²Alternatively, if y = u/v where u and v are function of x , then dy__ dx = v du__ dx − u dv__ dx/v²Example 10.12 Find dy__ dx if y = 1 + 3x/x²⁺1

Solution

y = 1 + 3x/x²⁺1 is a quotient with u = 1 + 3x and v = x²+ 1 du__ dx = 3 and dv__ dx = 2x Using the quotient rule: dy__ dx = v du__ dx − u dv__ dx/v²dy__ dx = 3(x²+ 1) − (1 + 3x)2x________________ ( x²+ 1)²= 3 x²+ 3 − 2x − 6x²______________ ( x²+ 1)²= 3 − 2x − 3x²_________ ( x²+ 1)²Example 10.13 Find dy__ dx if y = ²ˣ²_ ₓ − 2

Solution

y = 2x²_ x − 2 u = 2x²and v = x − 2 u′ = 4x and v′ = 1 dy__ dx = u′v − uv′_______ v²= (4x)(x − 2) − 2x²(1)______________ (x − 2)²= 4x²− 8x − 2x²___________ (x − 2)²= 2x²− 8x_______ (x − 2)²Example 10.14 Find dy__ dx if y = ˣ²_ √x + 1

Solution

u = x²and v = √x + 1 u′ = 2x and v = ( x + 1)¹__ 2 ⇒ v′ = 1/2( x + 1) −1/2 = 1/2√_____ x + 1 dy__ dx = u′v − uv′_______ v²= (√____ x + 1)(2x) − x²______ 2√____ x + 1________________ (√_ x + 1 ⁾²= (√____ x + 1)(2x) − x²______ 2√_____ x + 1________________ (x + 1) = 2(√x + 1) × √x + 1 × 2x − x²_____________________ 2 √x + 1 (x + 1) = 4x(x + 1) − x 2/2( x + 1)³_ ₂ dy__ dx = (3x + 4)_______ 2( x + 1)³_ ₂ Reciprocal Rule The reciprocal rule, which is derived from the quotient rule, is a tool for finding the derivative of a function that is a reciprocal to another function. If you have a function f(x) and you want to find the derivative of its reciprocal 1/f(x) ,we use the formula ( 1/f(x)) I = ᶠᴵ(x)_ (f(x) )²Derivation:

Let g(x) = 1/f(x) to find gᴵ(x), we can rewrite it as g(x) = ( f(x))−I ( f(x))−I : gᴵ(x) = − 1 × ( f(x))²× fᴵ(x) = − f I(x ) _ (f (x ) ) 2

Example 10.15

Find the derivative of h(x) = 1/x²

Solution

h(x) = 1/x²Step 1: Identify f(x) Here, f(x) = x²Step 2: Find fᴵ(x) f ᴵ(x) = d__ dx (x²) = 2x

Step 3: Apply to the reciprocal rule using the formula 1/f(x) = − ᶠᴵ(x)_ (f(x) )²hᴵ(x) = − x____ ( x²)²= −2x/x⁴= − 2/x 3

Example 10.16

Find the derivative of g(x) = 1______ 2x³+ 1

Solution

g(x) = 1______ 2x³+ 1

Step 1: Identify the function f(x) = 2x³+ 1

Step 2: Find the derivative of f(x) fᴵ(x) = d__ dx(2x³+ 1)) = 6x²Step 3: Apply the reciprocal rule 1/f(x) = − f I(x)_____ (f(x) )²Step 4: Substitute f(x) and fᴵ(x) into the formula f(x) = 2x³+ 1 and fᴵ(x) = 6x²gᴵ(x) = − f I(x)_____ (f(x) )²= − 6x 2________ ( 2x³+ 1 )²

Let us go through more examples that use the differentiation rules we have learnt.

Example 10.17

Product Rule:

Differentiate with respect to x, y = (x²+ 1)(x³+ 3)

Solution:

If y = uv, dy__ dx = v du__ dx + u dv__ dx u = x²+ 1 and v = x³+ 3 du_ dx = 2x, dv___ dx = 3 x²dy_ dx = 2x(x³+ 3) + 3 x²( x²+ 1)

Example 10.18

Quotient rule h(x) = x²_ x + 1

Solution

hᴵ(x) = u I v − u v I _ v 2 u(x) = x²and v(x) = x + 1 uᴵ(x) = 2x and vᴵ(x) = 1 Applying the quotient rule:

hᴵ(x) = 2x(x + 1) − x²___________ (x + 1)²hᴵ(x) = 2x²+ 2x − x²_ (x + 1)²hᴵ(x) = x²+ 2x_ (x + 1)²

Example 10.19

Chain rule:

Differentiate with respect tox, y = (x²+ 3)⁴Solution Chain rule: dy__ dx = y__ du × du__ dx Let u = x²+ 3, du__ dx = 2x, y = u⁴, y__ du = 4 u³dy_ dx = 4 u³× 2x = 8x (x²+ 3)³DIFFERENTIATING IMPLICIT FUNCTIONS The function which can be easily written as y = f(x) with the y variable on one side and the function of x on the other side, is called an explicit function. Some functions may not be given directly or explicitly. Examples include x²y − 3x = 2 and y⁴= 5x²y + 2xy and such functions are called implicit functions, where the x and the y variable cannot be written in the form y = f(x ). An implicit function has more than one solution for the given function.

Note the following where w . r . t x means “with respect to x ”.

1. When you differentiate y w . r . t x we obtain dy__ dx

2. When you differentiate y²w . r . t x we obtain 2y dy__ dx or d y²___ dx = 2y dy__ dx

3. When you differentiate y³w . r . t x we obtain d y³___ dx = 3y²dy__ dx

4. When you differentiate y⁴w . r . t x we obtain d y⁴___ dx = 4y³dy__ dx Also:

1. When you differentiate x²w . r . t x we obtain d x²___ dx = 2x

2. When you differentiate x³w . r . t x we obtain d x³___ dx = 3x²Example 10.20 Given that y³− 2 y²+ x²= 4, find dy___ dx .

Solution

y³− 2 y²+ x²= 4 3y²dy___ dx − 4y dy___ dx + 2x = 0 dy__ dx(3y²− 4y) = − 2x dy__ dx = 2x/3 y²− 4y

Example 10.21

Find dy___ dx , if 6y − 3 x⁴= 2 y⁶Solution 6y − 3 x⁴= 2 y⁶6 dy__ dx − 12 x³= 12 y⁵dy__ dx, divide through by 6 dy__ dx − 2 x³= 2 y⁵dy__ dx, collect your dy__ dxterms on one side and the remaining terms on the other dy__ dx − 2 y⁵dy__ dx = 2 x³dy__ dx(1 − 2y⁵) = 2 x³dy__ dx = x³______ 1 − 2y⁵Example 10.22 Find dy__ dx if x²+ y²− 2x + y = 6

Solution

x²+ y²− 2x + y = 6 2x + 2y dy___ dx − 2 + (1) dy___ dx = 0 2y dy___ dx + dy___ dx = 2 − 2x dy__ dx(2y + 1) = 2 − 2x dy__ dx = − 2x_____ 2y + 1 When differentiating the product of two functions, we apply the product rule.

• If y = uv , dy__ dx = v du__ dx + u dv__ dx We use this when we do implicit differentiation.

For example, the derivative of x²y³with respect to x, u = x², du__ dx = 2x and v = y³, dv__ dx = 3 y²dy__ dx ∴ d( x²y³)_______ dx = 2xy³+ 3x²y²dy___ dx

Example 10.23

Find dy__ dx if x³+ y³= 3xy

Solution

Differentiate (w. r. t. x) term by term y³as a composite function and 3xy as a product.

Then 3x²+ 3y²dy__ dx = 3y + 3x dy__ dx 3y²dy__ dx − 3x dy__ dx = 3y − 3x²dy__ dx(3y²− 3x) = 3y − 3 x²dy__ dx = 3y − 3 x²_______ 3y²− 3x dy__ dx = y − x²_____ y²− x

Example 10.24

Find dy__ dx if , x²y − 5x = 3

Solution

x²y − 5x = 3 We differentiate with respect to x (w.r.t x) term by term throughout.

To differentiate x²y consider it as uv and use the product rule.

x²dy__ dx + y × 2x v du__ dx + u dv__ dx , so we have by differentiating x²y − 5x = 3 x²dy__ dx + 2xy − 5 = 0 Now find dy__ dx by making it the subject of the formula.

x²dy___ dx = 5 − 2xy dy_ dx = 5− 2xy/x²Example 10.25 Find dy__ dx if 4y²x − 5 x²y²+ 4y = 0

Solution

4y²x − 5 x²y²+ 4y = 0 Remember 4y²x and − 5 x²y²are products, so we must use the product rule:

4y²+ 4x × 2y dy__ dx − 10x y²− 10 x²y dy__ dx + 4 dy__ dx = 0 4y²+ 8xy dy__ dx − 10x y²− 10 x²y dy__ dx + 4 dy__ dx = 0 Isolate dy__ dx terms and simplify:

(8xy − 10x²y + 4) dy___ dx = 10x y²− 4 y²( 4xy − 5x²y + 2) dy__ dx = y²(5x − 2) dy_ dx = y²(5x − 2)____________ 4xy − 5 x²y + 2 Everyday problems involving derivatives

Example 10.26

Two cars start from the same point. Car X travels north at 40mph and car Y travels east at 30 mph. After 2 hours, how fast is the distance between the two cars changing.

Solution

x : Distance Car Y has travelled east y : Distance Car X has travelled north D : Distance between the two cars Express distance after t hours x = 30t and y = 40t For the relationship between distances use Pythagoras’ theorem;

D²= x²+ y²Differentiate implicitly with respect to time (t ) 2D dD___ dt = 2x dx__ dt + 2y dy__ dt (Simplify by dividing through by 2) D dD___ dt = x dx___ dt + y dy___ dt After 2 hours:

x = 30 × 2 = 60 miles y = 40 × 2 = 80 miles dx__ dt = 30mph and dy__ dt = 40 mph Solving for D D²= x²+ y²D²= (60)²+ (80)²D²= 3600 + 6400 D²= 10000 D = 100 Substitute into the related rate equation D dD___ dt = x dx__ dt + y dy__ dt 100 dD___ dt = 60 × 30 + 80 × 40 100 dD___ dt = 1800 + 3200 100 dD___ dt = 5000 dD___ dt = 5000/100 = 50mph .

Therefore, the rate of change of the distance with respect to time after 2 hours is 50 mph.

Example 10.27

A cylindrical tank has a radius of 2 metres and is filled with water at a rate of 3 cubic metres per minute. How fast is the height the water rising when the water is 4 metres deep?

Solution

1. Cylinder Dimensions: Radius(r) = 2 metres

2. Volume of Rate (water is being filled) = dV___ dt = 3m³/ min.

3. Height of water (we need to find the rate of change with water height ( dh__ dt ) when the height (h) is 4 metres.

Steps:

1. The volume of a cylinder is given by V = πr²h

2. Differentiate with respect to time (t) ⇒ dV___ dt = πr²× dh__ dt

3. Substitute known values: dV___ dt = 3m³/ min and r = 2m ∴ 3 = π × (2)²× dh__ dt

4. Solve for dh__ dt ⇒ 3 = 4 π × dh__ dt dh__ dt = 3__ π Using π ≈ 3.14 dh__ dt = 3/4 × 3.14 = 3/12.56 = 0.239m / min.

Therefore, the rate at which the water is rising when 4m deep is 0.24m per minute (to 2sf).

Example 10.28

A conical tank has a height of 6m and a base radius of 3m. Water is pumped into the tank at a rate of 2 cubic metres per minute.

How fast is the water level rising when the water is 2m deep?

Solution

1. Conical tank dimension:

Height (H) = 6m and base radius (R) = 3m

2. Volume Rate:

Water is being pumped in at dV___ dt = 2m³/ min

3. Height of water:

We need to find the rate of change of the water height, ( dh__ dt ), when the height (h) is 2m Steps:

1. Volume of the water:

The volume of cone is given by V = 1/3 π r²h Where r and h are the radius and height of the water. Since the water forms a smaller, similar cone within the tank, we have:

r/R = h/H ⇒ r = R/H h = 3/6 h = h/2 Substitute r = h/2 into the volume formula:

V = 1/3 × π × ( h/2 )²× h V=1/3 × π × h/4³= πh³___ 12

2. Differentiate with respect to time (t):

dV___ dt = π__ 12 × 3 h²× dh__ dt = π h²___ 4 × dh__ dt

3. Substitute known values:

dV___ dt = 2m³/ min and h = 2m dV___ dt = π h²___ 4 × dh__ dt 2 = π( 2)²____ 4 × dh___ dt Simplify:

2 = π × 4/4 × dh___ dt 2 = π × dh___ dt dh__ dt = 2__ π Using π ≈ 3.14 dh__ dt = 2/3.14 = 0.636 m / min Therefore, the water level is rising at 0.64m/min (to 2sf) when h = 2m.

Transcendental functions are those that are not algebraic, meaning they cannot be expressed as a finite combination of the basic algebraic operations (addition, subtraction, division, multiplication, taking square roots etc). They include trigonometric, exponential and logarithmic functions.

Common Transcendental Functions:

1. Exponential Functions : f(x) = eˣThese involve the constant e (approximately 2.71828). This is the base of natural logarithms.

2. Logarithmic Functions: f(x) = ln(x) The natural logarithm function is the inverse of the exponential function.

3. Trigonometric Functions:

a. Sine: f(x) = sin(x)

b. Cosine: f(x) = cos(x)

c. Tangent: f(x) = tan(x) Derivative of Exponential and Logarithmic Functions The process of differentiating transcendental functions follows specific rules:

1. Exponential Functions:

d__ dx eˣ= eˣ(The derivative of eˣis itself) If g(x) = eᶠ(x) then d__ dx(g(x)) = f′(x) × eᶠ(x)

2. Logarithmic Functions:

d__ dx ln(x) = 1/x (The derivative of the natural logarithm ln(x) is 1/x .

If f(x) = logₐ(g(x)) then d__ dx fx = 1____ ln(a) × 1/g(x) × g′(x)

Example 10.29

Find the derivative of the following

1. f(x) = e²ˣ2. y = e⁻³ˣ3. g(t) = e²ᵗ²+t

4. Differentiate the function y = xe⁻²ˣSolution

1. f′(x) = d(2x)____ dx × e²ˣ= 2e²ˣ2. dy__ dx = d( − 3x)______ dx × e⁻³ˣ= − 3e⁻³ˣ3. g′(t) = d(2t²+ t)_______ dt × e²ᵗ²+t = (4t + 1) e²ᵗ²+t

4. y = xe⁻²ˣUsing the product rule followed by the chain rule:

dy__ dx = x d (e⁻²ˣ)______ dx + e⁻²ˣd(x)___ dx = x e⁻²ˣd( − 2x)______ dx + e⁻²ˣ= − 2x e⁻²ˣ+ e⁻²ˣ= e⁻²ˣ(1 − 2x)

Example 10.30

Find the derivative of y = ³√lnx

Solution

y = ( ln(x))¹__ 3 dy__ dx = 1/3 (ln(x))−2/3 × 1/x = 1__ 3x × 1________ ₃ √(ln(x))²= 1________ 3x ³√(ln(x))²Derivative of trigonometric functions Trigonometric Functions:

• Sine: d__ dx sin(x) = cos(x)

• Cosine: d__ dx cosx = − sin(x)

• Tangent: d__ dx tan(x) = sec²(x) Differentiation of sinx and cosx from first principles Let f(x) = sinx f(x + h) = sin(x + h) (where x is in radians) f(x + h) − f(x)__________ h = sin(x + h) − sinx/h Using the trigonometry identity:

sinA − sinB = 2cos 1/2(A + B)sin 1/2(A − B) When A = x + h and B = x, we have sin(x + h) − sinx = 2cos 1/2(x + h + x)sin 1/2(x + h − x) = 2cos(x + 1/2 h)sin( 1/2 h) sin(x + h) − Sinx/h = 2cos(x + 1/2 h)Sin( 1/2 h) ________________ h = cos(x + 1/2 h) × 2sin( 1__ 2h) _______ h limₕ→0 sinx + h − sinx/h = limₕ→0 [cos(x + 1/2 h) × 2sin (1/2 x) _____ 2(1/2 h)]

Note: limₕ→0 (x + 1/2 h) = cosx and limₕ→0 sin( 1/2 x)_____ 1/2 x = 1 This shows that if y = sinx, then dy__ dx = cosx or d(sinx)_____ dx = cosx Let f(x) = cosx f(x + h) = cos(x + h) f( x + h) − f(x)___________ h = cos( x + h) − cosx/h Using the trigonometry identity:

cosA − cosB = − 2sin1_ 2(A + B)sin1_ 2(A − B).

When A = x + h and B = x, we have;

cos(x + h) − cosx = − 2sin1/2(x + h + x)sin1/2(x + h − x) = − 2sin(x + 1/2 h)sin( 1/2 h) cos(x + h) − cosx/h = 2sin(x + 1/2 h)sin( 1__ 2h)_________________ h = − sin(x + 1/2 h) × 2sin(1/2 h) _______ h limₕ→0 cos(x + h) − cosx/h = limₕ→0 [ − sin(x + 1/2 h) × 2sin( 1 _ 2 h)__________ 2 ( 1/2 h) ] limₕ→0 − sin(x + 1/2 h) × limₕ→0 sin( 1/2 h)______ 1/2 h

Note: limₕ→0 sin(x + 1/2 h) = sinx and limₕ→0 sin( 1 _ 2 x)_________ 1/2 x = 1 − sinx × 1 = − sinx That shows that if y = cosx, then dy__ dx = − sinx or d(cosx)______ dx = − sinx.

Example 10.31

Differentiate with respect to x, sin3x.

Solution

Let y = sin3x and let u = 3x Then y = sinu dy__ du = cosu and du__ dx = 3 Using the chain rule: dy__ dx = y__ du × du__ dx dy__ dx = cosu × 3 = 3cosu, but u = 3x dy__ dx = 3cos3x

Example 10.32

If y = sin(2x − 4), find dy__ dx.

Solution

y = sin(2x − 4), let u = 2x − 4 , then y = sinu dy__ du = cosu and du__ dx = 2 Using the chain rule: dy__ dx = y__ du × du__ dx dy__ dx = cosu × 2 = 2cosu, but u = 2x − 4 dy__ dx = 2cos(2x − 4)

Example 10.33

If y = cos( ^(π)__ 4 − 2x), find dy__ dx

Solution

y = cos( ^(π)__ 4 − 2x), let u = π__ 4 − 2x , then y = cosu dy__ du = − sinu and du__ dx = − 2 Using the chain rule: dy__ dx = y__ du × du__ dx dy__ dx = − sinu × − 2 = 2sinu , but u = π__ 4 − 2x dy__ dx = 2sin( π__ 4 − 2x)

Example 10.34

If y = sin x⁰, find dy__ dx

Solution

We can only differentiate if the angle is in radians, so we must convert the x degrees to radians, x⁰= π__ 80 x y = sin x⁰y = sin π__ 80 x dy__ dx = cos π__ 80 x × π__ 80 π___ 180 cos x⁰Note that the result is NOT Cos x⁰. It must be stressed that the formula for differentiating trigonometric functions is only true if the angles are expressed in radians.

Remember that angles in radian measure are written simply as x, θ, 3x, etc. Angles in degree measure must have the degree symbol (°).

Example 10.35

Given that y = sin(2x), find dy__ dx

Solution

y = sin(2x) Let u = 2x ⇒ du__ dx = 2 y = sinu ⇒ d y_ dx = cosu By the chain rule, dy__ dx = dy__ du × du__ dx dy__ dx = cosu × 2 = 2cosu , but u = 2x ∴ dy___ dx = 2cos2x

Example 10.36

Find the derivative of the function Sinx/x .

Solution

y = sinx/x is a quotient:

Let u = sinx ⇒ du__ dx = cosx v = x ⇒ dv__ dx = 1 Using the quotient rule, dy__ dx = v du _ dx − u dv__ dx/v²:

dy__ dx = xcosx − sinx_________ ( x)²Example 10.37 Given that y = cosx____ √x , find dy__ dx

Solution

y = cosx____ √x is a quotient.

Let u = cosx ⇒ du__ dx = − sinx V = √x = x ¹__ 2 ⇒ dv___ dx = 1/2 x−1/2 Using the quotient rule: dy__ dx = v du _ dx − u dv__ dx/v²dy__ dx = x¹__ 2 (− sinx) − cosx × ( 1/2 x–1/2 ) ___________________ (√_ x)²= −√x sinx − cosx/2 √x/x = − 2 √x × √x sinx − cosx/2 √x/x dy__ dx = − 2 x sinx − cosx____________ 2x √x

Example 10.38

Given that y = x²sin(x), find dy__ dx

Solution

y = x²sin(x) is a product.

u = x², du___ dx = 2x v = sinx, dv___ dx = cosx Using the product rule, dy__ dx = v du__ dx + u dv__ dx :

dy_ dx = sinx × 2x + x²× cosx = 2xsinx + x²cosx

1. Find dy__ dx if y = x²+ 5√x

2. Find dy__ dx if y = x⁴− 2x + 1/x²3. Find dy__ dx if y = 3 √x

4. Find dy__ dx if y = 3__ √_ x

5. Differentiate with respect to x , ( 2x − 1)²6. Differentiate with respect to x , (2x + 4)(3x − 1)

7. Find dy__ dx if 3x⁴+ 2 x²− 1___________ 2x²8. Differentiate with respect to x 1_______ √4x − 7

9. Find dy__ dx if y = x²(2x − 5)⁴10. Differentiate y = (3x − 1)³( x²+ 5)

11. Find dy__ dx if y = ( x²+ 1)¹__ 2

12. Find dy__ dx if y = 4_______ √_____ 1 − 2x

13. If y⁴+ x⁴− 2x²y²= 9, find dy__ dx

14. Differentiate with respect to x, tan(4x − 1)

15. Find dy__ dx ,if xcosy + ycosx = 2

16. If y 5_______ (1 – x²)³, show that (1 − x²) dy__ dx = 6xy

17. Differentiate with respect to x, y = x²cosx

18. Differentiate with respect to x, y = xsinx z

Welcome to this fascinating section on integration! In this section, we will uncover the basics of integration, a fundamental concept in mathematics that plays a vital role in various fields. Integration allows us to find areas, volumes and other quantities that accumulate over time, making it essential for understanding the world around us.

To fully appreciate integration, we will build on your existing knowledge of differentiation. Differentiation helps us understand how things change; like the speed of a car or the growth of a plant. By studying these changes, we can make predictions and decisions in our daily lives. Integration and differentiation are closely linked. While differentiation focuses on rates of change, integration helps us find the total accumulation of those changes. In physics, when we differentiate the position of an object over time, we find its velocity. Conversely, by integrating velocity, we can determine the total distance travelled. In economics, differentiation helps us understand how supply and demand change over time, while integration allows us to calculate total revenue or costs over a period. Also, Engineers use integration to design structures and analyse forces, in biology, integration helps model population growth and resource consumption and Environmental Scientists use it to calculate the total amount of resources consumed or pollutants produced over time.

KEY IDEAS

• An antiderivative is a fundamental concept in calculus that is closely related to integration. It is essentially the reverse process of differentiation.

• Calculating the area of rectangles is fundamental in understanding the approximation of areas under curves

• Definite Integrals represent the total accumulation of a quantity over a specific interval. It is written as ∫ₐ ᵇf(x)dx where f(x) is the function being integrated and a and b are the limits of integration.

• Indefinite Integrals represent a family of functions whose derivative is the integrand. It is written as : ∫ f(x)dx = F(x) + C where F(x) is the antiderivative of f(x), and C is the constant of integration.

• Partitioning intervals involve dividing a given interval on the number line into smaller subintervals of equal or varying lengths. This concept is fundamental in calculus, particularly in approximation techniques like integration.

Activity 11.1: Partitioning an area within an interval Working in small groups, carry out the following activity.

A frog is to hop into a pond, as shown below. Study the number line and answer the questions that follow in your groups.

Figure 11.1: Number line

1. What is the interval from the starting point of the frog to the pond?

2. How many steps was the frog hopping from one point to the other?

3. How many steps in total did the frog hop to the pond?

4. Now perform the following task:

a. Draw a number line with the interval [ 0, 18].

b. If you move three steps each, how many sub-divisions will you have?

c. Discuss how the sub-divisions can be found in the absence of a number line.

Understanding Intervals

An interval [a, b] represents all the numbers between a and b, including the endpoints. For example, the interval [1, 5] includes all numbers from 1 to 5, inclusive.

Introducing Partitioning

Partitioning an interval means dividing it into smaller, manageable parts called subintervals. This is particularly useful for approximating areas under curves.

Notation for Partitioning

Let x₀, x₁ , x₂, … .. xₙ be the points that define the partition, where:

x₀ = a (the left endpoint of the interval), xₙ =b (the right endpoint of the interval) The points in between represent the divisions of the interval into subintervals.

For example, with n=4 for the interval [1, 5] the points would be: 1, 2, 3, 4, 5, so x₀ = 1, x₁ = 2, x 4 = 3, x₃ = 4, x₄ = 5 .

Calculating the width, Δx:

To find the width of each subinterval in the interval [a, b]:

The width of each subinterval is given by: Δx = b − a/n where a is the a is the left endpoint of the interval and b is the right endpoint of the interval For our example with a = 1 and b = 5, and n = 4: Δx = 5 − 1/4 This means each subinterval has a width of 1.

Subintervals:

Define the subintervals as:

The first subinterval is [x₀, x₁] = [1, 2] The second subinterval is [x₁, x₂] = [2, 3] The third subinterval is [x₂, x₃] = [3, 4] The fourth subinterval is [x₃, x₄] = [4, 5] Importance of Partitioning: Partitioning an interval allows us to approximate the area under a curve by breaking it down into simpler shapes (rectangles) over these subintervals.

Example 11.1

Determine the subintervals for the interval [0, 15] for a step size of 3?

Solution

Step size of 3 means we add 3 to the left endpoint which is 0 until we get the right endpoint which is 15:

0 + 3 = 3 3 + 3 = 6 6 + 3 = 9 9 + 3 = 12 12 + 3 = 15 The first subinterval is [0,3] The second subinterval is [3,6] The third subinterval is [6,9] The fourth subinterval is [9,12] The fifth subinterval is [12,15] There are 5 subintervals.

Example 11.2

Determine the subintervals for the interval [0, 6] for a step size of 0.5

Solution

The step size of 0.5 means we add 0.5 to the left endpoint which is 0 until we get to the endpoint of 6.

x 0 0.5 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5 5.5 6.0 x+0.5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 We have [0, 0.5], [0.5, 1.0], [1.0, 1.5], [1.5, 2.0], [2.0, 2.5], [2.5, 3.0], [3.0, 3.5], [3.5, 4.0], [4.0, 4.5], [4.5, 5.0], [5.0, 5.5], [5.5, 6.0], There are 12 sub intervals.

Example 11.3

Find the width for 4 subintervals for interval [1, 2], hence write the subintervals.

Solution

The width of each subinterval (step size) is given by:

Δx = b − a/n where a is the a is the left endpoint of the interval and b is the right endpoint of the interval a=1 and b=2, and n=4:

Δx = 2 − 1/4 = 0.25 Subintervals = 1, 1 + 0.25 = 1.25, 1.25 + 0.25 = 1.5, 1.5 + 0.25 = 1.75, 1.75 + 0.25 = 2 1, 1.25, 1.5, 1.75, 2

Example 11.4

Find the width for 5 subintervals for interval [0, 1], hence write the subintervals.

Solution

The width of each subinterval (step size) is given by:

Δx = b − a/n where a is the left endpoint of the interval and b is the right endpoint of the interval a = 0 and b = 1, and n = 5:

Δx = 1 − 0/5 = 0.2 Subintervals = 0, 0 + 0.2 = 0.2, 0.2 + 0.2 = 0.4, 0.4 + 0.2 = 0.6, 0.6 + 0.2 = 0.8, 0.8 + 0.2 = 1, 0, 0.2, 0.4, 0.6, 0.8, 1

Example 11.5

Find the width for 10 subintervals for interval [-2, 3], hence write the subintervals.

Solution

The width of each subinterval is given by:

Δx = b − a/n where a is the left endpoint of the interval and b is the right endpoint of the interval a=-2 and b=3, and n = 10:

Δx = 3 − ( − 2)________ 10 = 0.5 Subintervals: -2, -1.5, -1.0, -0.5, 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0

Example 11.6

Find the width for 9 subintervals for interval [1, 4], hence write the subintervals

Solution

The width of each subinterval is given by:

Δx = b − a/n where a is the left endpoint of the interval and b is the right endpoint of the interval a= 1 and b=4, and n = 9:

Δx = 4 − (1)______ 9 = 1/3 Subintervals: 1, 4/3, 5/3, 2, 7/3, 8/3, 3, 10/3 , 11/3 , 4

Activity 11.2: Approximating the area under a curve Working in pairs, or individually, carry out the following activity.

Figure 11.2: Graph of quadrilateral ABCD

1. What type of quadrilateral is ABCD

2. Find the area of the quadrilateral.

3. Is it a good approximation for the area under the curve from x = -1 to x = 5?

4. How could the approximation be improved?

Example 11.7

Find the area of the trapeziums A₁to A₄to estimate the area under the curve. Will this be an over or underestimation? How can our estimate be improved?

Figure 11.3: Graph of trapeziums A₁to A₄

Solution

a b h Area = 1/2(a + b)h A₁ 2.4 4 2 1_ 2(2.4 + 4)(2) = 6.4 A₂ 4 4.6 2 1_ 2(4 + 4.6)(2) = 8.6 A₃ 4.6 4 2 1_ 2(4.6 + 4)(2) = 10.6 A₄ 4 2.2 2 1_ 2(4 + 2.2)(2) = 6.2 The approximated area is A₁+ A₂+ A₃+ A₄= 6.4 + 8.6 + 10.6 + 6.2 = 31.8 squa red un i ts.

This will be an underestimation as we are not including the area immediately below the curve and the top of the trapeziums.

To improve our estimation, we could increase the number of trapeziums by increasing the number of partitions.

Example 11.8

Find the area of the function y = 2x + 2 as shown in the diagram below, by calculating the area of the individual rectangles and adding them together.

Is this an over or underestimation of the actual area?

Figure 11.4: Graph of the function y = 2x+2

Solution

Rectangle a h Area = ah 1 4 1 4 x 1 = 4 2 6 1 6 x 1 = 6 3 8 1 8 x 1 = 8 4 10 1 10 x 1 = 10 5 12 1 12 x 1 = 12 6 14 1 14 x 1 = 14 7 16 1 16 x 1 = 16 The approximated area is A₁+ A₂+ A₃+ A₄+ A₅+ A₆+ A₇= 4 + 6 + 8 + 10 + 12 + 14 + 16 = 70 squa red un i ts.

This is an overestimation of the area as we have the counted the additional blue triangles in our total.

Introduction to Partitioning

We have learned that when we want to find the area under a curve within a specific interval, using more partitions (trapeziums or rectangles) gives us a better approximation of the actual area. As we increase the number of partitions, the approximated area gets closer to the exact area.

Step-by-Step Computation

To approximate the area under a curve f(x) from x = a to x = b using rectangles.

Number of Rectangles (= number of subintervals) We divide the interval [a, b] into n equal parts (subintervals).

The width of each rectangle is denoted as h.

Calculating Width

The width, h, can be calculated using the formula: h = b− a/n For example, if a = 0 and b = 10 with n = 5, then: h = 10 − 0/5 = 2 Finding the x Coordinates If we choose to use the right side of the rectangles, the x coordinates of the right edges of the rectangles are:

x₁ = a + h x₂ = a + 2h x₃ = a + 3h xₙ = a + nh For our example:

x₁ = 0 + 2 = 2 x₂ = 0 + 2(2) = 4 x₃ = 0 + 3(2) = 6 x₄ = 0 + 4(2) = 8 x₅ = 0 + 5(2) = 10 Finding the Heights:

The heights of the rectangles are given by the function values at these x coordinates:

f(x₁) = f(a + h) f(x₂) = f(a + 2h) f(x₃) = f(a + 3h) f(xₙ) = f(b) = f(a + nh) Calculating the Area of Each Rectangle:

The area of each rectangle can be calculated as:

A₁ = h × f( x₁) A₂ = h × f( x₂) A₃ = h × f( x₃) Aₙ = h × f( xₙ) Total Approximate Area To find the total approximate area under the curve, we sum the areas of all rectangles:

R(n) = A₁ + A₂ + A₃ + … .. Aₙ This can be expressed as: R(n)= h × ∑ ᵢ₌₁ ⁿf(a + ih) Or as a definite integral:

If we want to find the area under the curve from x=a to x=b, we write ∫ₐ ᵇf(x)dx where a is the lower limit and b is the upper limit.

Approximation Using Definite Integrals:

The area can be approximated as:

Area ∫ₐ ᵇf(x) ≈ [f(x₁) ∆ x + f(x₂) ∆ x + f(x₃) ∆ x…… ..f(xₙ) ∆ x] Where Δx is the width of each subinterval.

Example 11.10

Find the definite integral of ∫₀ ⁴(x²+ 2)dx on the interval [0, 4], using 8 subintervals.

Solution

h = b− a/n a = 0, b = 4, n = 8 h = 4 − 0/8 h = 0.5 x 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 f(x) = (x²+ 2 ) 2.25 3 4.25 6 8.25 11 14.25 18 Using, R(n)= h × ∑ ᵢ₌₁ ⁿf(a + ih), where h= 0.5, ∑ ᵢ₌₁ ⁴f(a + ih) = [2.25 + 3 + 4.25 + 6 + 8.25 + 11 + 14.25 + 18] = 67 R(n)= 0.5 × 67 = 33.5 square units Alternatively:

The area can be approximated as:

Area ∫ₐ ᵇf(x) ≈ [f(x₁) ∆ x + f(x₂) ∆ x + f(x₃) ∆ x…… ..f(xₙ) ∆ x] = [2.25(0.5) + 3(0.5) + 4.25(0.5) + 6(0.5) + 8.25(0.5) + 11(0.5) + 14.25(0.5) + 18 (0.5 )] = 33.5 square units Thus, the approximate value of the definite integral ∫₀ ⁴(x²+ 2)dx, using 8 subintervals ≈ 33.5 square units.

Example 11.11

Find the approximate value of ∫₁ ⁹( 1 __ x 2 )dx using 5 subintervals

Solution

h = b− a/n a = 1, b = 9, n = 5 h = 9− 1/5 h = 1.6 x 2.6 4.2 5.8 7.4 9 f(x) = 1/x²0.1479 0.0566 0.0297 0.0182 0.0123 Using, R(n)= h × ∑ ᵢ₌₁ ⁿf(a + ih), where h= 1.6, ∑ ᵢ_(=2.6) ⁹f(a + ih) = [0.1479 + 0.0566 + 0.0297 + 0.0182 + 0.0123] = 0.2467 R(n)= 1.6 × 0.2647 0.4235 square units Alternatively:

The area can be approximated as:

Area ∫ₐ ᵇf(x) ≈ [f(x₁) ∆ x + f(x₂) ∆ x + f(x₃) ∆ x…… ..f(xₙ) ∆ x] = [0.1479(1.6) + 0.0566(1.6) + 0.0297(1.6) + 0.0182(1.6) + 0.0123(1.6)] = [ 0.23664 + 0.09056 + 0.04752 + 0.02912 + 0.01968] = 0.4235 square units

Example 11.12

Find the approximate value of ∫₁ ³√1 + 2 x²dx using 6 subintervals. Give your answer to 4 significant figures.

Solution

h = b− a/n a = 1, b = 3, n = 6 h = 3− 1/6 h = 1 __ x 4_ 3 5_ 3 2 7_ 3 8_ 3 3 f(x) = √1 + 2 x²√41_ 3 √59_ 3 3 √107_ 3 √137_ 3 √19 Using, R(n)= h × ∑ ᵢ₌₁ ⁿf(a + ih), where h = 1/3 ∑ ᵢ_(=2.6) ⁹f(a + ih) = [√__ 41/3 + √__ 59/3 + 3 + √___ 107/3 + √___ 137/3 + √__ 19] = 19.4032 R(n)= 1/3 × 19.4032 = 6.468 square units (to 4sf).

Activity 11.3: Revision of Differentiation

In groups, pairs or individually solve the following questions and then share your results with the class.

1. Differentiate the following functions listed in the Table 11.1.

Table 11.1: Function and its derivative Function Derivative

a. 2 x³+ 6

b. 2 x³+ 8

c. − 3 x⁴+ 2.5

d. − 3 x³+ 16

e. 2 x³+ 5x + 3

f. 2 x³+ 5x − 8

g. x⁵− 5 x²+ 4

h. x⁵− 5 x²− 9

2. Once you have completed the differentiation, look at your results. Identify any pairs of functions that have the same derivative. Write them down.

3. Discuss in your groups what made the functions with the same derivatives different from each other?

Hint: Consider aspects such as constants added or subtracted from the functions.

Observations

• The functions 3x²− 8 and 3 x²+ 7 have the same derivative 6x . The only difference is the constant (− 8 and 7).

• Similarly, 2 x³+ 6 and 2 x³+ 3 share the derivative. The only difference is the constant (6 and 3).

This means that if we are given the derivatives to find the function, we must make sure that we make provision for a constant term. The process of finding the function f( x ) when we know the derivative f’(x) is called Integration or Antiderivative.

An antiderivative of a function is the reverse process of finding a derivative.

While differentiation helps us determine how a function changes, finding an antiderivative allows us to reconstruct the original function from its rate of change.

If f′(x) is the derivative of a function f(x), then f(x) is called an antiderivative of f′(x). This relationship can be expressed mathematically as:

∫f′(x)dx = f(x) + c where c is the constant of integration. This constant is important because there are infinitely many antiderivatives for any given derivative, all differing by a constant as shown in Activity 11.3.

Finding the antiderivatives Given f( x ) = xⁿ, to find f’( x ) we:

i. Multiplied the function by the power, n

ii. Subtracted 1 from the power, n – 1.

If integration is the reverse of differentiation, then study Table 11.2 for steps in finding the antiderivative.

Table 11.2: Steps in finding the antiderivative Differentiation Integration (Reverse of Differentiation)

Step 1: Multiply the function by the power, n n xⁿStep 1: Add 1 to the power n, to obtain n+1 xⁿ⁺¹Step 2: Subtract 1 from the power, n n xⁿ⁻¹Step 2: Divide the function by the new power n+1 xⁿ⁺¹_ n + 1 NB: Add the constant of integration, c, to all indefinite integrals.

Indefinite Integrals are when we have to find the antiderivative without limits.

Indefinite Integrals

Example 11.13

Integrate ∫(x²+ 4x)dx

Solution

To integrate we add one to the power and divide by the new power, so we have:

∫(x²+ 4x)dx = x²⁺¹____ 2 + 1 + 4x¹⁺¹_____ 1 + 1 + c = x³__ 3 + 4x²___ 2 + c = x³__ 3 + 2 x²+ c

Example 11.14

Simplify ∫(3x³− 18 x²+ 5x)dx

Solution

∫(3x³− 18 x²+ 5x)dx = 3x³⁺¹_____ 3 + 1 − 18x²⁺¹_____ 2 + 1 + 5x¹⁺¹_____ 2 + c = 3 x⁴___ 4 − 18x³____ 3 + 5x²___ 2 + c = 3x⁴___ 4 − 6x³+ 5x²___ 2 + c

Example 11.15

∫ 4dx

Solution

4 can be written as 4 x⁰∫ 4 x⁰dx = 4 x⁰⁺¹_____ 0 + 1 + c = 4 x¹___ 1 + c = 4x + c This shows how a constant term in a given function is integrated.

Example 11.16

Find the integral of ∫ (4x⁴+ 3 x²+ 1/x³+ 6)dx

Solution

∫ (4x⁴+ 3 x²+ 1/x³+ 6)dx = ∫ (4x⁴+ 3 x²+ x⁻³+ 6)dx = 4x⁵___ 5 + x³− 1/2 x²+ 6x + c

Example 11.17

Find the integral of ∫ x⁵− 3 x³+ 8x − 13/x³dx

Solution

Simplify each term in the numerator with the term in the denominator x⁵__ x³− 3x³___ x³+ 8x/x³− 13/x³= x²− 3 + 8 x⁻²− 13 x⁻³∫(x²− 3 + 8 x⁻²− 13 x⁻³)dx = x³__ 3 − 3x − 8/x + 13/2 x²+ c

Suppose f(x) is a continuous function over the interval [a, b]. We can define:

∫ₐ ᵇf(x)dx = F(b) − F(a) where F is any antiderivative of f(x).

The Fundamental Theorem of Calculus shows a fascinating relationship between differentiation and integration. In simple terms, it tells us that the process of finding an integral can be thought of as the reverse of finding a derivative.

Let us take the function f(x) = x³as an example to illustrate this concept.

Step 1: Finding the Derivative

To find the derivative f′(x), we use the power rule. The power rule states that if you have a function of the form xⁿ, the derivative is calculated by:

a. Reducing the exponent by 1.

b. Multiplying the resulting expression by the original exponent.

So, for our function f(x) = x³, f′(x) = 3 x²Step 2: Finding the Integral Now, let’s find the integral of f(x). To do this, we will reverse the process we used for differentiation:

a. Increase the exponent by 1.

b. Divide by the new exponent.

For the function f(x) = x³:

F(x)=∫f(x) dx = ∫x³dx.

Following the steps:

a. Increase the exponent: 3 + 1 = 4

b. Divide by the new exponent F(x) = x⁴__ 4 + c, where c is the constant of integration.

The expression, ∫ₐ ᵇf(x)dx is a definite integral because the limits are known i.e., the upper limit is b and the lower limit is a. However,∫ f(x)dx is indefinite integral because the limit is not known Let us work through the following steps to evaluate a Definite Integral

1. Write down the integral you want to evaluate, e.g., ∫ₐ ᵇf(x)dx, where a and b are the limits of integration.

2. Determine the antiderivative F(x) of the function f(x). This involves using basic integration techniques

3. Compute F(b) and F(a):

a. F(b): Substitute the upper limit b into the antiderivative.

b. F(a): Substitute the lower limit a into the antiderivative.

4. The value of the definite integral is given by: ∫ₐ ᵇf(x)dx = F(b)−F(a).

Example 11.18

Evaluate ∫₁ ⁴(3x²+ 2x + 6)dx

Solution

∫₁ ⁴(3x²+ 2x + 6)dx Finding the Antiderivative:

F(x) =[x³+ x²+ 6x]4 1 Substitute the limits = [(4)³+ (4)²+ 6(4)] − [ (1)³+ (1)²+ 6(1) = [64 + 16 + 24] − [1 + 1 + 6] = [104] − [8] = 96

Example 11.19

Evaluate ∫₋₃ ⁰(t³+ 5 t²− 5)dt

Solution

Finding the Antiderivative:

F(x) = [t⁴__ 4 + 5t³__ 3 − 5t] 0 − 3 Substitute the limits:

= [ (0)⁴____ 4 + 5(0)³____ 3 − 5(0)] − [(− 3)⁴____ 4 + 5(− 3)³_____ 3 − 5( − 3)] = [0] − [81/4 − 135/3 + 15] = [0] − [− 117/12 ] = 117/12 = 39/4

Example 11.20

Evaluate ∫₁ ²(x + 3)(x − 5)dx

Solution

∫₁ ²( x²− 2x − 15)dx = [x³__ 3 − x²− 15x]2 1 Substitute the limits:

[(2)³___ 3 − (2)²− 15(2)] − [(1)³___ 3 − (1)²− 15(1)] = [8/3 − 4 − 30] − [1/3 − 1 − 15] = [− 94/3 ] − [− 47/3 ] = − 47/3

Activity 11.4: Fundamental Theorem of Calculus

Working in pairs, carry out the following activity.

1. The graph below represents the function f(x) = 2x on the interval [0, 4]

a. Find the area of the figure ABC.

b. Find the integral of f(x) = 2x on the interval [0, 4]

Figure 11.5: Graph of the function f(x) = 2x

2. Repeat the steps above for the graph f(x) = 2x + 3 on the interval [-1, 2] Compare your results for (1) and (2). Look at the graph below to help with this.

Does the result of the integral match the area under the line?

Figure 11.6: Graph of f(x) = 2x + 3 Therefore, we can use the idea of the Fundamental Theorem of Calculus to help us calculate the area under curves accurately.

Example 11.21

Evaluate the following, remembering that what you are finding is the area between the curve and the x axis between the given limits.

Note, if the answer is negative, this means the curve, and its corresponding area, is below the x-axis.

1. ∫₁ ³( 2x²+ 5x + 2)dx

2. ∫₋₂ ⁰( 5x³+ 4 x²− 3x − 8)dx

3. ∫₂ ⁴(x⁴– x³__ 2 – 4x + 9)dx

4. ∫√2 2 (2x − 1)dx

Solution

1. ∫₁ ³( 2x²+ 5x + 2)dx F(x) = [ 2/3 x³+ 5 x²___ 2 + 2x]3 1 Substitute the limits:

[2/3 (3)³+ 5 (3)²____ 2 + 2(3)] − [2/3 (1)³+ 5 (1)²____ 2 + 2(1)] = [18 + 45/2 + 6] − [2/3 + 5/2 + 2] = [93/2 ] − [31/6 ] = 248/6 = 124/3

2. ∫₋₂ ⁰( 5x³+ 4 x²− 3x − 8)dx = [ 5/4 x⁴+ 4___ 3x³− 3/2 x²− 8x] 0 − 2 =[5/4 (0)⁴+ 4/3 (0)³−3/2 (0)²− 8(0)]− [5/4 ( − 2)⁴+ 4/3 (−2)³− 3/2 (− 2)²− 8(−2)] = [0] − [5/4(16) + 4/3(− 8) − 3/2(4) + 16)] = [0] − [20 − 32/3 − 6 + 16] = [0] − [20 − 32/3 − 6 + 16] = 0 − [58_ 3 ] = − 58/3

3. ∫₂ ⁴( x⁴− x³__ 2 − 4x + 9)dx = [x⁵__ 5 − x⁴__ 8 − 2 x²+ 9x]4 2 = [ (4)⁵____ 5 − (4)⁴___ 8 − 2 (4)²+ 9(4)] − [(2)⁵___ 5 − (2)⁴___ 8 − 2 (2)²+ 9(2)] = [1024/5 − 256/8 − 32 + 36] − [32/5 − 16/8 − 8 + 18] = [884_ 5 ] − [72_ 5 ] = 812/5

4. ∫√2 2 (2x − 1)dx = [x²− x] 2 √2 = [(2)²− (2)] − [( √2)²− √2] = [4 − 2] − [2 − √2] = [2] − [2 − √2] = √2

1. Given the interval [0, 20] and a step size of 4, find the various intervals and hence find the number of sub-intervals.

2. There are 8 subintervals with a step size of 1.5. If the interval is [1, t] find the value of t.

3. Find the approximated area under the curve below.

.

Figure 11.7

4. Find the approximate value of ∫₁ ³( x²+ 1)dx using 8 subintervals.

5. Evaluate ∫₀ ⁸1/x²+ 1dx using 4 subintervals.

6. Evaluate y = ∫₀ ⁶(x³+ 2x²− 6)dx

7. Integrate the following:

a. − 5___ 9x⁴b. 8x

c. 6 v²+ 3 √v/v

d. 8x³− 5 √x − 6

e. 10 − 1___ √x + 1/x⁴

f. 3x⁴+ 16x_______ 4x³8. A businessman found that the rate of change in the cost of GH¢, of producing x thousands of calculators is given by dy__ dx = 1500/x²and that the overhead cost is GH¢ 10 000.00.

9. Find the cost function, y, and hence find the cost of producing 500 calculators, leaving your answer correct to the nearest Ghanaian Cedi.

10. Given that dy__ dx = 3x²+ 4x + 5 and y = 20 when x = 2, find the value of y when x = 13.

11. The rate of change of an area(A) with respect to the radius is given by dA___ dr = 5r + 12 .

If A = 24 when r = 2, find r when A = 152.

12. A curve is such that dy__ dx = 3x²+ 10x + m , where m is a constant and that it passes through the points (2,6) and (4,18). Find the equation of the curve.

13. Evaluate:

a. ∫₀ ¹(2x⁹+ x⁴+ 1)dx

b. ∫₂ ⁴(2 + x²)dx

c. ∫₋₁ ²(3 t²− 4)dt

d. ∫₁ ⁴1/y³dy

e. ∫₋₃ ³(4 w²− 4w)dw

f. ∫₁ ⁵1__ √t dt