Principles of Calculus

This topic builds upon the concepts you learned in your second year and is a fundamental aspect of calculus. Integration is a branch of calculus that supports us in finding the approximate values of functions. It is used to find the area under curves, marginal cost, etc. It is also a prerequisite for engineering science, physics, etc. In this section, we will look at the integration of functions, including logarithms and natural logarithms. We shall also look at how to approximate the area under a curve using the trapezium rule.

KEY IDEAS

• Antiderivatives : If f′(x) represents the derivative of a function $f(x)$ , then $f(x)$ is known as an antiderivative of $f'(x)$ . This is expressed as $\int f'(x)\,dx = f(x) + C.$ • Finding Antiderivatives : For $f(x) = x^n$ o Differentiate: $\frac{d}{dx}(x^n) = nx^{n-1}$ o Integrate: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$ (where $n \neq -1$ ). • Methods of Substitution : When an integral can be expressed in the form $\int f[g(x)]g'(x)\,dx$ , use $u = g(x)$ to simplify the integral. • Transcendental Functions : Functions like $e^x$ and $\ln(x)$ that cannot be expressed as algebraic equations. Integration examples include $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ and $\int \ln|x|\,dx = x\ln|x| + C$ .

Understanding Antiderivatives

If $f'(x)$ represents the derivative of a function $f(x)$ , then $f(x)$ is known as an antiderivative of $f'(x)$ . This relationship can be mathematically expressed as: $\int f'(x)\,dx = f(x) + c$ , where $c$ is the constant of integration. This constant is crucial because there are infinitely many antiderivatives corresponding to any given derivative, each differing by a constant.

Finding Antiderivatives

For the function $f(x) = x^n$ : Recall the following steps:

1. Differentiation: Multiply the function by the power n and subtract 1 from the power to obtain: $nx^{n-1}$
2. Integration (Reverse of Differentiation): a. Add 1 to the power n to get $n + 1 : x^{n+1}$ b. Divide the function by the new power $\frac{x^{n+1}}{n+1} + c$

Activity 5.1 Understanding Antiderivatives through Integration

Concept Overview When we integrate the derivative of a function $f'(x),$ we obtain the original function $f(x)$ plus a constant: $\int f'(x)\,dx = f(x) + c$ For example, if $f(x) = x^3$ $\frac{dy}{dy} = 3x^2$ $f'(x) = 3x^2$ How will change or reverse the result, $\frac{dy}{dy} = 3x^2$ back to $f(x) = x^3$ ? The procedure of reversing $f'(x) = 3x^2$ to $f(x) = x^3$ is using Integration denoted by the symbol ∫ Integrating Power Functions

1. Given the function : $f(x) = x^n$
2. Integration Steps : Step 1: Add 1 to the exponent n to obtain $n + 1$ Step 2: Divide by the new exponent: $\int x^n\,dx =$ $\frac{x^{n+1}}{n+1} + c$ Where c is the constant of integration . Indefinite Integral is when we have to integrate a function which has no limits. For example, $\int x^4\,dx$ . On the other hand, $\int_1^2 x^4\,dx$ , is a definite integral, because it has a limit [1, 2]. Other rules of integration are as follows

Rule Function Procedure Power rule $f(x) = ax^n$ $\int f(x)\,dx = \int ax^n\,dx$ = $\frac{ax^{n+1}}{n + 1} + c$ , $n \neq -1$ Sum rule $f(x) + h(x)$ $\int [f(x) + h(x)]\,dx = \int f(x)\,dx + \int h(x)\,dx$ Difference rule $f(x) - h(x)$ $\int [f(x) - h(x)]\,dx = \int f(x)\,dx - \int h(x)\,dx$ Multiplication rule $kf(x)$ $\int kf(x)\,dx = k\int f(x)\,dx$

Example 5.1

Identify the various rules that will be applicable in evaluating the following

1. $\int 5x^3\,dx$
2. $\int 6x\,dx$
3. $\int (2x - 8x^2)\,dx$
4. $\int (3x^3 + 4x^2)\,dx$
5. $\int \left(\frac{x^4}{x^2} + 4\right)dx$

Solution

1. The power rule and the multiplication rule
2. The power rule
3. The difference rule, the multiplication rule
4. The sum rule and the multiplication rule
5. The sum rule

Definite Integral

Definite integrals are integrals that have upper and lower limits and can be evaluated as a numerical value. The Second Fundamental Theorem of Calculus is a key concept in calculus that connects two important ideas: definite integrals and antiderivatives Definite Integral : The definite integral $\int_a^b f(x)\,dx$ represents the area under the curve of the function $f(x)$ from $x = a$ to $x = b$ . It helps us find the total accumulation of a quantity (like area, distance, etc.) over an interval.

Antiderivative : An antiderivative of a function $f(x)$ is another function $F(x)$ such that when you take the derivative of $F(x),$ you get back $f(x).$ In other words, $F'(x) = f(x).$ For example, if $f(x) = 2x$ , then an antiderivative $F(x)$ could be $x^2$ , because the derivative of $x^2$ is $2x$ . The Second Fundamental Theorem of Calculus states: $\boldsymbol{\int_a^b f(x)\,dx = F(b) - F(a)}$ This means that to find the area under the curve $f(x)$ from $a$ to $b$ , we:

1. Find an Antiderivative $F(x)$ of $f(x).$
2. Evaluate $F(x)$ at the endpoints b and a.
3. Subtract the value at a from the value at $b$ : $F(b) - F(a).$ The Second Fundamental Theorem of Calculus simplifies the process of finding the area under a curve by allowing us to use antiderivatives instead of relying solely on approximations. It shows the powerful relationship between differentiation and integration, making calculus a more manageable and effective tool for solving real-world problems.

Example 5.2

Evaluate the following: a. $\int_{-1}^{2} (x^2 + 2x)\,dx$ b. $\int_0^3 (6 - 3x + 2x^2)\,dx$ c. $\int_{-2}^{4} 5\,dx$ d. $\int_0^\pi (\sin x)\,dx$

Solution

a. $\int_{-1}^{2} (x^2 + 2x)\,dx$ Find an Antiderivative: $F(x)$ as $\left[\frac{x^3}{3} + x^2\right]_{-1}^{2}$ Evaluate: $F(x)$ at the endpoints 2 and -1 as $F(2) =$ $\left[\frac{2^3}{3} + (2)^2\right]$ $F(-1) =$ $\left[\frac{(-1)^3}{3} + (-1)^2\right]$ Subtract the value at a from the value at b: $F(2) - F(-1).$ $F(2) - F(1)$ = $\left[\frac{8}{3} + 4\right] - \left[\frac{-1}{3} + 1\right]$ $= \left[\frac{20}{3}\right] - \left[\frac{2}{3}\right]$ $\int_{-1}^{2} (x^2 + 2x)\,dx =$ 6

b. $\int_0^3 (6 - 3x + 2x^2)\,dx$ Find an Antiderivative $F(x):$ as $\left[6x - \frac{3x^2}{2} + \frac{2x^3}{3}\right]_0^3$ Evaluate $F(x)$ at the endpoints 3 and 0 as; $F(3) = \left[6(3) - \frac{3(3)^2}{2} + \frac{2(3)^3}{3}\right]$ $F(0) = \left[6(0) - \frac{3(0)^2}{2} + \frac{2(0)^3}{3}\right]$ Subtract the value at a from the value at b: $F(3) - F(0).$ $F(3) - F(0) =$ $\left[18 - \frac{27}{2} + 18\right] - [0]$ $= \left[\frac{45}{2}\right] - [0]$ $\int_0^3 (6 - 3x + 2x^2)\,dx = \frac{45}{2}$ c. $\int_{-2}^{4} 5\,dx = [5x]_{-2}^{4}$ $= [5(4)] - [5(-2)]$ $= [20] - [-10]$ $= 30$ Hence $\int_{-2}^{4} 5\,dx = 30$ d. $\int_0^\pi (\sin x)\,dx = [-\cos x]_0^\pi$ $= [-\cos(\pi)] - [-\cos(0)]$ $= [-(-1)] - [-1]$ $= [1 + 1]$ $= [2]$ $\int_0^\pi (\sin x)\,dx = 2$

Methods of Substitution

If an integral can be written in the form $\int f[g(x)]g'(x)\,dx,$ then we can set $u = g(x)$ and integrate $\int f(u)\,du$ . In other words, when an integral is not in a standard algebraic or trigonometric form, it can often be simplified using substitution.

Example 5.3

Let us go through this process step-by-step using the integral $\int \left(\sqrt{2x + 4}\right)dx$ as an example. Step 1: Choose a Substitution Let $u = 2x + 4$ Step 2: Differentiate the Substitution. Next, we differentiate u with respect to $x$ : $\frac{du}{dx} = 2$ Step 3: Solve for $dx$ or make $dx$ the subject. Now, we rearrange this equation to express dx in terms of du: $dx = \frac{1}{2}du$ Step 4: Substitute into the Integral. Now we substitute u and dx into the original integral. The integral becomes $\int \left(\sqrt{2x + 4}\right)dx = \int \sqrt{u} \times \frac{1}{2}du$ Step 5: Simplify the Integral This can be rewritten as: $\frac{1}{2}\int (\sqrt{u})\,du$

Example 5.4

In each of the following, rewrite the integral in the form $\int f[g(x)]g'(x)\,dx$ . Identify $g(x)$ and $g'(x)$ .

1. $\int \cos(2x)\,dx$
2. $\int x^2\sqrt{3x^3 + 4}\,dx$
3. $\int 4t^3(-6t^4 + 10)\,dt$
4. $\int \frac{5x^3}{\sqrt{3 + 10x^4}}\,dx$

Solution

1. Let $u = g(x) = 2x$ $\frac{du}{dx} = g'(x) = 2$ $du = 2\,dx$ $dx = \frac{du}{2}$ $\int \cos(2x)\,dx = \int \cos(u) \times \frac{du}{2}$ $= \frac{1}{2}\int \cos(u)\,du$

Now, $\int \cos(2x)\,dx = \frac{1}{2}\int \cos u\,du$ 2. $\int x^2\sqrt{3x^3 + 4}\,dx$ Let $u = g(x) = 3x^3 + 4$ $\frac{du}{dx} = g'(x) = 9x^2$ $du = 9x^2\,dx$ $dx = \frac{du}{9x^2}$ $\int x^2(\sqrt{u}) \times \frac{du}{9x^2}$ $\frac{1}{9}\int (\sqrt{u})\,du$ Now, $\int x^2\sqrt{3x^3 + 4}\,dx = \frac{1}{9}\int (\sqrt{u})\,du$ 3. $\int 4t^3(-6t^4 + 10)\,dt$ Let $u = g(t) = -6t^4 + 10$ $\frac{du}{dt} = g'(t) = -24t^3$ $du = -24t^3\,dt$ $dt = \frac{du}{-24t^3}$ $\int 4t^3(-6t^4 + 10)\,dt = \int 4t^3(u) \times \frac{du}{-24t^3}$ $= -\frac{1}{6}\int (u)\,du$ Now, $\int 4t^3(-6t^4 + 10)\,dt = -\frac{1}{6}\int (u)\,du$ 4. $\int \frac{5x^3}{\sqrt{3 + 10x^4}}\,dx$ Let $u = g(x) = 3 + 10x^4$ $\frac{du}{dx} = g'(x) = 40x^3$ $du = 40x^3\,dx$ $dx = \frac{du}{40x^3}$ $\int \frac{5x^3}{\sqrt{3 + 10x^4}}\,dx = \int \frac{5x^3}{\sqrt{u}} \times \frac{du}{40x^3}$ $= \frac{1}{8}\int \frac{1}{\sqrt{u}}\,du$ Now, $\int \frac{5x^3}{\sqrt{3 + 10x^4}}\,dx = \frac{1}{8}\int \frac{1}{\sqrt{u}}\,du$

Activity 5.2

Working in small groups, evaluate the following indefinite integral using the substitution method. $\int \cos(4x + 3)\,dx$ Step 1: Choose a Substitution Let: $u = 4x + 3$ Step 2: Differentiate the Substitution Differentiate $u$ with respect to $x$ : $\frac{du}{dx} = 4$ Step 3: Solve for $dx$ or make $dx$ the subject Express dx in terms of $du$ : $dx = \frac{1}{4}du$ Step 4: Substitute into the Integral Substitute u and dx into the original integral. The integral becomes: $\int \cos(4x + 3)\,dx = \int \cos u \times \frac{1}{4}du$ Step 5: Simplify the Integral This can be rewritten as: $\frac{1}{4}\int \cos u\,du$ Step 6: Integrate $\frac{1}{4}\int \cos u\,du =$ $\frac{1}{4}[\sin u] + C =$ $\frac{1}{4}\sin u + C$ Step 7: Substitute Back Finally, we substitute back $u = 4x + 3$ to obtain $\frac{1}{4}\sin(4x + 3) + C$ . Thus, the integral $\int \cos(4x + 3)\,dx$ evaluates to $\frac{1}{4}\sin(4x + 3) + C$ .

Example 5.5

Evaluate $\int 2xe^{x^2}\,dx$

Solution

Let $u = x^2$ $\frac{du}{dx} = 2x$

$dx = \frac{1}{2x}du$ $\int 2xe^{x^2}\,dx = \int 2xe^u \times \frac{1}{2x}du$ $\int 2xe^{x^2}\,dx = \int e^u\,du$ $= e^u + c$ Thus $\int 2xe^{x^2}\,dx =$ $e^{x^2} + c$

Example 5.6

Simplify $\int x^3\cos x^4\,dx$

Solution

Let $u = x^4$ $\frac{du}{dx} = 4x^3$ $dx = \frac{1}{4x^3}du$ $\int x^3\cos x^4\,dx = \int x^3\cos u \times \frac{1}{4x^3}du$ $\int x^3\cos x^4\,dx = \frac{1}{4}\int \cos u\,du$ $= \frac{1}{4}(\sin u) + c$ $= \frac{1}{4}\sin(x^4) + c$

Example 5.7

Evaluate $\int x^2(x^3 + 5)^9\,dx$

Solution

Let $u = x^3 + 5$ $\frac{du}{dx} = 3x^2$ $dx = \frac{1}{3x^2}du$

$\int x^2(x^3 + 5)^9\,dx = \int x^2(u)^9 \times \frac{1}{3x^2}du$ $\int x^2(x^3 + 5)^9\,dx = \frac{1}{3}\int u^9\,du$ $= \frac{1}{3}\left[\frac{u^{10}}{10}\right] + C$ $= \frac{u^{10}}{30} + C$ $= \frac{(x^3 + 5)^{10}}{30} + C$ The procedure for evaluating a definite integral follows the same steps as we have witnessed above. However, we must change the original given limits to conform with $g(x) = u$ .

Example 5.8

Evaluate $\int_0^1 x\sqrt{x^2 + 2}\,dx$ Step 1: Choose a Substitution Let’s define a new variable u based on the expression inside the square root: Let $u = x^2 + 2$ Step 2: Change the Limits of Integration Next, we need to change the limits of integration according to our substitution: When $x = 1$ , $u = 1^2 + 2$ $u = 3$ At $x = 0$ $u = 0^2 + 2$ $u = 2$ Step 3: Find $du$ Now, we need to find the differential $du$ : $u = x^2 + 2$ $\frac{du}{dx} = 2x$ $du = 2x\,dx$ $dx = \frac{du}{2x}$

Step 4: Substitute into the Integral Now, we substitute u and dx into the integral: $\int_2^3 x\sqrt{u} \times \frac{du}{2x}$ $\frac{1}{2}\int_2^3 \sqrt{u}\,du$ Step 5: Rewrite the Integral Rewrite the integral as $\frac{1}{2}\int_2^3 u^{\frac{1}{2}}\,du$ Step 6: Integrate Now, we can integrate: $\frac{1}{2}\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_2^3$ $\frac{1}{3}\left[u^{\frac{3}{2}}\right]_2^3$ Step 7: Evaluate the Limits Now we evaluate the limits: $\frac{1}{3}\left(\left[3^{\frac{3}{2}}\right] - \left[2^{\frac{3}{2}}\right]\right)$ $\frac{1}{3}(3\sqrt{3} - 2\sqrt{2})$ Working in pairs, or small groups, solve the following example.

Example 5.9

Evaluate $\int_{-1}^{0} x^3(4x^4 + 2)^3\,dx$

Solution

Let $u = 4x^4 + 2$ $\frac{du}{dx} = 16x^3$ $dx = \frac{du}{16x^3}$ When $x = -1,$ $u = 4(-1)^4 + 2$ $u = 6$

When $x = 0$ , $u = 4(0)^4 + 2$ $u = 2$ $\int_{-1}^{0} x^3(4x^4 + 2)^3\,dx = \int_6^2 x^3(u)^3\frac{du}{16x^3}$ $\int_{-1}^{0} x^3(4x^4 + 2)^3\,dx = \frac{1}{16}\int_6^2 u^3\,du$ $= \frac{1}{16}\left[\frac{u^4}{4}\right]_6^2$ $= \frac{1}{16}\left(\left[\frac{(2)^4}{4}\right] - \left[\frac{(6)^4}{4}\right]\right)$ $= \frac{1}{16}\left(\left[\frac{16}{4}\right] - \left[\frac{1296}{4}\right]\right)$ $= \frac{1}{16}([4] - [324])$ $= -20$ Therefore, $\int_{-1}^{0} x^3(4x^4 + 2)^3 = -20$ .

Indefinite integrals

What is an Indefinite Integral?

• An indefinite integral is an integral that does not have specific limits (i.e., no starting and ending points). • It is written as $\int f(x)\,dx$ and represents a family of functions that are all antiderivatives of $f(x)$ . • Since there are many antiderivatives for a function (they can differ by a constant), we include an arbitrary constant $C$ to represent all possible antiderivatives.

Example 5.10

Evaluate the following a. $\int (x^6 + 2)\,dx$ b. $\int \sqrt{(x)}\,dx$ c. $\int (2x^2 + 3x + 4)\,dx$ d. $\int (e^{2x} + 5x)\,dx$

Solution

a. $\int (x^6 + 2)\,dx = \frac{x^{6+1}}{6+1} + \frac{2x^{0+1}}{0+1} + c$ $= \frac{x^7}{7} + 2x + c$ d. $\int \sqrt{(x)}\,dx = \int (x)^{\frac{1}{2}}\,dx$ $= \frac{(x)^{\frac{1}{2}+1}}{\frac{1}{2} + 1} + c$ $= \frac{(x)^{\frac{3}{2}}}{\frac{3}{2}} + C$ $= \frac{2}{3}x^{\frac{3}{2}} + C$ e. $\int (2x^2 + 3x + 4)\,dx = \frac{2x^{2+1}}{2+1} + \frac{3x^{1+1}}{1+1} + \frac{4x^{1+0}}{1+0} + C$ $= \frac{2x^3}{3} + \frac{3x^2}{2} + 4x + C$ f. $\int (e^{2x} + 5x)\,dx = \frac{1}{2}e^{2x} + \frac{5x^2}{2} + C$ Integration can be applied to solve real-life problems involving marginal cost, velocity, distance and acceleration. Some terms and symbols are represented in the table below. Quantity Definition Notation Displacement Distance from a fixed origin $s, y, x$ Velocity Rate of change of displacement with time $v = \frac{ds}{dt}$ Acceleration Rate of change of velocity with time $a = \frac{dv}{dt}$

Example 5.11

A particle $Q$ moves in a straight line so that at time $t$ , seconds, its acceleration is $(12t + 3)\mathrm{ms}^{-2}$ . $Q$ passes through the point $O$ with velocity 5m $s^{-1}$ . Find the: a. velocity of $Q$ in terms of $t$ b. distance $Q$ from $O$ when $t = 3\mathrm{s}$

Solution

a. The velocity of Q in terms of $t$ , acceleration $a = \frac{dv}{dt}$ $\int dv = \int a\,dt$ $v = \int a\,dt$ $v = \int (12t + 3)\,dt$ $v = 6t^2 + 3t + C$ But at $t = 0$ , the velocity was $5\mathrm{ms}^{-1}$ $5 = 6(0)^2 + 3(0) + C$ $C = 5$ Hence, the velocity of Q in terms of t is $v = 6t^2 + 3t + 5$ b. $v = \frac{ds}{dt}$ $ds = \int v\,dt$ $s = \int v\,dt$ $s = \int (6t^2 + 3t + 5)\,dt$ $s = 2t^3 + \frac{3}{2}t^2 + 5t + C$ At the point $O, s = 0, t = 0$ $0 = 2(0)^3 + \frac{3}{2}(0)^2 + 5(0) + C$ $C = 0$ $s = 2t^3 + \frac{3}{2}t^2 + 5t$ At $t = 3\mathrm{s}$ , $s = 2(3)^3 + \frac{3}{2}(3)^2 + 5(3)$ $s = 54 + \frac{27}{2} + 15$ $82.5m$

Activity 5.3 Exploring Transcendental Functions

This activity is to enable you to research, understand, and present information about transcendental functions, their properties, and applications. Working in small groups:

1. Research the properties, graphs, and applications of exponential functions, including growth and decay models.

2. Investigate the properties, graphs, and real-world applications of logarithmic functions, including their role in solving exponential equations.

3. Explore the properties and applications of trigonometric functions, including their use in modelling periodic phenomena.

4. Research inverse trigonometric functions, their properties, and applications in geometry and calculus.

5. Investigate hyperbolic functions, their definitions, properties, and applications in engineering and physics.

6. Make use of reliable online resources such as educational websites, academic articles, and videos (e.g., Khan Academy ( khanacademy.org ), Wolfram Alpha ( wolframalpha.com ), Math is Fun ( mathsisfun.com ) or university resources).

7. Take notes on key points. Transcendental functions are special types of functions that cannot be expressed as solutions to simple algebraic equations (like polynomials). In other words, you cannot write them as a finite combination of addition, subtraction, multiplication, division, and taking roots.

Example of Algebraic Functions

Example of Algebraic Functions

For example, consider the equation: $y^2 - 2x = 0.$

You can solve this equation for $y$ to get: $y = \mp\sqrt{2x}$ This means $y$ is an algebraic function because we can express it using a finite number of operations.

Example of Transcendental Functions

Now, let's look at a transcendental function like: $y = e^x$ . This function represents the exponential function, which grows very rapidly. There is no polynomial equation that can describe it simply. Other examples of transcendental functions include trigonometric functions (like sine and cosine) and natural logarithms (like ln(x)). Here we will concentrate on the integration of exponential functions as well as natural logarithms. Figure 5.1 represents an algebraic function ( $y = x^2)$ and transcendental functions ( $y = e^x)$ . What is your observation? [Figure] Figure 5.1: Graphical Illustration of $y = x^2$ and $y = e^x$

Example 5.12

Evaluate

Solution

Applying $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ We have: $\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C$ Alternatively, we can consider:

Let $u = 3x$ $\frac{du}{dx} = 3$ $du = 3\,dx$ $dx = \frac{du}{3}$ $\int e^u\frac{du}{3} =$ $\frac{1}{3}\int e^u\,du =$ $\frac{1}{3}e^u + C =$ $\frac{1}{3}e^{3x} + C$

Example 5.13

E valuate $\int e^{2-7x}\,dx$

Solution

Let $u = 2 - 7x$ $\frac{du}{dx} = -7$ $du = -7\,dx$ $dx = -\frac{du}{7}$ $\int -e^u\frac{du}{7} =$ $-\frac{1}{7}\int e^u\,du =$ $-\frac{1}{7}e^u + C =$ $-\frac{1}{7}e^{2-7x} + C$

Example 5.14

Evaluate $\int e^{2x}\sqrt{1 + e^{2x}}\,dx$

Solution

Let $u = 1 + e^{2x}$ $\frac{du}{dx} = 2e^{2x}$ $du = 2e^{2x}\,dx$ $dx = \frac{du}{2e^{2x}}$ $\int e^{2x}(\sqrt{u}) \times \frac{du}{2e^{2x}} =$ $\frac{1}{2}\int (\sqrt{u})\,du =$ $\frac{1}{2}\int u^{\frac{1}{2}}\,du =$ $\frac{1}{2}\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right] + C =$ $\frac{1}{3}u^{\frac{3}{2}} + C =$ $\frac{1}{3}(1 + e^{2x})^{\frac{3}{2}} + C$

Example 5.15

Evaluate $\int 4x^2e^{2x^3}\,dx$

Solution

Let $u = 2x^3$ $\frac{du}{dx} = 6x^2$ $du = 6x^2\,dx$ $dx = \frac{du}{6x^2}$ $\int 4x^2(e^u) \times \frac{du}{6x^2} =$ $\frac{2}{3}\int (e^u)\,du =$ $\frac{2}{3}[e^u] + C =$ $\frac{2}{3}\left(e^{2x^3}\right) + C =$ $\frac{2e^{2x^3}}{3} + C$

Example 5.16

Evaluate $\int e^x(3e^x - 2)^2\,dx$

Solution

Let $u = 3e^x - 2$ $\frac{du}{dx} = 3e^x$ $dx = \frac{du}{3e^x}$ $\int e^x(u)^2 \times \frac{du}{3e^x}$ $\frac{1}{3}\int (u)^2\,du =$ $\frac{1}{3}\left[\frac{u^3}{3}\right] + C =$ $\frac{u^3}{9} + C =$ $\frac{(3e^x - 2)^3}{9} + C$

Example 5.17

Evaluate $\int_1^2 e^{1-2x}\,dx$

Solution

Let $u = 1 - 2x$ $\frac{du}{dx} = -2$ $dx = -\frac{du}{2}$ When $x = 1$ $u = 1 - 2(1)$ $u = -1$

When $x = 2$ $u = 1 - 2(2)$ $u = -3$ $\int_1^2 e^{1-2x}\,dx = \int_{-1}^{-3} e^u \times -\frac{du}{2}$ $= -\frac{1}{2}\int_{-1}^{-3} e^u\,du$ $= -\frac{1}{2}[e^u]_{-1}^{-3}$ $= -\frac{1}{2}([e^{-3} - e^{-1}])$ $= -\frac{1}{2}(0.367879 - 0.0497871)$ $= 0.1590$

Example 5.18

$\int_0^1 x\left(e^{4x^2+1} - 8\right)dx$

Solution

Let $u = 4x^2 + 1$ $\frac{du}{dx} = 8x$ $dx = \frac{du}{8x}$ When $x = 0$ $u = 4(0)^2 + 1$ $u = 1$ When $x = 1$ $u = 4(1)^2 + 1$ $u = 5$ $\int_0^1 x\left(e^{4x^2+1} - 8\right)dx = \int_1^5 x(e^u - 8)\frac{du}{8x}$ $= \frac{1}{8}\int_1^5 (e^u - 8)\,du$ $= \frac{1}{8}[e^u - 8u]_1^5$ $= \frac{1}{8}([e^5 - 8(5)] - [e^1 - 8(1)])$ $= \frac{1}{8}([e^5 - 40] - [e^1 - 8])$ $= \frac{1}{8}([148.413 - 40] - [2.71828 - 8])$ $= \frac{1}{8}([108.413] - [-5.28172])$ $= \frac{1}{8}(113.69472)$ $= 14.21184$ Thus, $\int_0^1 x\left(e^{4x^2+1} - 8\right)dx = 14.2118$ (approximated to four decimal places).

Integrating natural logarithmic functions

The integral of a natural logarithmic function in the form Integrand Brief explanation The absolute value is important because the logarithm is $\int \frac{1}{x}\,dx =$ $\ln|x| + C$ only defined for positive numbers. By using $|x|,$ we can handle both positive and negative values of $x$ .

This integral shows that if you have a function u and its $\int \frac{u'}{u}\,dx = \ln|u| + C$ derivative natural logarithm $u'$ , you can of $u$ integrate . the fraction $\frac{u'}{u}$ to get the $u$ can be any differentiable function. The integral of the natural logarithm function $\ln|x|$ results in a more complex expression involving $x$ multiplied by $\ln|x|$ , minus $x$ , plus a constant C. $\int \ln|x|\,dx = x\ln|x| - x + C$ NB: This integral is often solved using a technique called integration by parts, which is based on the formula: $\int u\,dv = uv - \int v\,du$ .

Example 5.19

Evaluate the following a. $\int \frac{2}{x}\,dx$ b. $\int \frac{1}{4x - 1}\,dx$ c. $\int \frac{7}{3 - x}\,dx$ d. $\int \frac{x^2 - 4}{x}\,dx$ e. $\int \frac{3x^2 + 1}{x^3 + x}\,dx$ f. $\int \frac{\sec^2 x}{\tan x}\,dx$ g. $\int \frac{x}{x^2 + 2}\,dx$

Solution

a. $\int \frac{2}{x}\,dx = 2\int \frac{1}{x}\,dx$ $= 2\ln|x| + C$ b. $\int \frac{1}{4x - 1}\,dx$ Let $u = 4x - 1$ $\frac{du}{dx} = 4$ $du = 4\,dx$ $dx = \frac{1}{4}du$ $\int \frac{1}{4x - 1}\,dx = \int \frac{1}{u} \times \frac{1}{4}du$ $= \frac{1}{4}\int \frac{1}{u}\,du$ $= \frac{1}{4}\ln|u| + C$ $= \frac{1}{4}\ln|4x - 1| + C$

Thus $\int \frac{1}{4x - 1}\,dx = \frac{1}{4}\ln|4x - 1| + C$ c. $\int \frac{7}{3 - x}\,dx$ Let $u = 3 - x$ $\frac{du}{dx} = -1$ $du = -dx$ $dx = -du$ $\int \frac{7}{3 - x}\,dx = \int \frac{7}{u} \times -du$ $= -7\int \frac{1}{u}\,du$ $= -7\ln|u| + C$ $= -7\ln|3 - x| + C$ d. $\int \left(\frac{x^2 - 4}{x}\right)dx$ Divide each term by $x$ $\int \left(\frac{x^2}{x} - \frac{4}{x}\right)dx = \int \left(x - \frac{4}{x}\right)dx$ $= \int x\,dx - \int \frac{4}{x}\,dx$ $= \frac{x^2}{2} - \ln|x| + C$ e. $\int \frac{3x^2 + 1}{x^3 + x}\,dx$ Let $u = x^3 + x$ $du = (3x^2 + 1)\,dx$ $dx = \frac{du}{3x^2 + 1}$ $\int \frac{3x^2 + 1}{x^3 + x}\,dx = \int \frac{3x^2 + 1}{u} \times \frac{du}{3x^2 + 1}$ $= \int \frac{\cancel{3x^2 + 1}}{u} \times \frac{du}{\cancel{3x^2 + 1}}$ $= \int \frac{1}{u}\,du$ $= \ln|u| + C$ $= \ln|x^3 + x| + C$ Thus $\int \frac{3x^2 + 1}{x^3 + x}\,dx$ = $\ln|x^3 + x| + C$ . f. $\int \frac{\sec^2 x}{\tan x}\,dx$ Let $u = \tan x$ $\frac{du}{dx} = \sec^2 x$ $dx = \frac{du}{\sec^2 x}$ $\int \frac{\sec^2 x}{\tan x}\,dx = \int \frac{\sec^2 x}{u} \times \frac{du}{\sec^2 x}$ $= \int \frac{1}{u}\,du$ $= \ln|u| + C$ $= \ln|\tan x| + C$ Thus $\int \frac{3x^2 + 1}{x^3 + x}\,dx$ = $\ln|\tan x| + C$ . g. $\int \frac{x}{x^2 + 2}\,dx$ Let $u = x^2 + 2$ $\frac{du}{dx} = 2x$

$dx = \frac{du}{2x}$ $\int \frac{x}{x^2 + 2}\,dx = \int \frac{x}{2x\,(u)}\,du$ $= \frac{1}{2}\int \frac{1}{u}\,du$ $= \frac{1}{2}\ln|u| + C$ Thus $\int \frac{x}{x^2 + 2}\,dx$ $= \frac{1}{2}\ln|u| + C$