AOS4 Topic 4: Integration

Introduction to Integration

The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. In a broad sense, in calculus, the idea of limit is used where algebra and geometry are implemented. Limits help us in the study of the result of points on a graph such as how they get closer to each other until their distance is almost zero.

Maths Integration

In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Calculation of small addition problems is an easy task which we can do manually or by using calculators as well. But for big addition problems, where the limits could reach to even infinity, integration methods are used. Integration and differentiation both are important parts of calculus.

Integration – Inverse Process of Differentiation

Differentiation is the process of finding the derivative of the functions and integration is the process of finding the antiderivative of a function. So, these processes are inverse of each other. The integration is also called the anti-differentiation. In this process, we are provided with the derivative of a function and asked to find out the function

Definition

Integration is the process of calculating a function's integral, which represents the function's accumulated sum over a particular time frame. To determine the total value, infinitely many tiny values are added together.


Notation:

The integral of a function \( f(x) \) with respect to \( x \) over an interval \([a, b]\) is denoted by:

\[ \int_{a}^{b} f(x) \, dx \]

Types of Integrals

  1. Indefinite Integral:

    The indefinite integral, also known as the antiderivative, of a function \( f(x) \) is another function \( F(x) \) whose derivative is equal to the original function \( f(x) \). Mathematically, it is denoted as:

    \[ \int f(x) \, dx = F(x) + C \]

    where:

    • \( f(x) \) is the integrand, representing the function to be integrated.
    • \( F(x) \) is the antiderivative or indefinite integral of \( f(x) \).
    • \( dx \) represents the variable of integration.
    • \( C \) is the constant of integration, representing the family of functions that differ from each other by a constant.

    We have two functions: \( f(x) = x^2 \) and its antiderivative \( F(x) = \frac{1}{3}x^3 \). The function \( f(x) \) represents a quadratic curve, while \( F(x) \) represents the accumulation of areas under the curve of \( f(x) \).

    The antiderivative \( F(x) \) of \( f(x) \) is obtained by integrating \( f(x) \) with respect to \( x \). Mathematically, we have:

    \[ F(x) = \int f(x) \, dx = \int x^2 \, dx = \frac{1}{3}x^3 + C \]

    where \( C \) is the constant of integration. Here, we've taken \( C = 0 \) for simplicity.

    In the graph, the blue curve represents \( f(x) = x^2 \), while the orange curve represents its antiderivative \( F(x) = \frac{1}{3}x^3 \). As we can see, the antiderivative \( F(x) \) is the curve that results from finding the area under the curve of \( f(x) \) and accumulating it with respect to \( x \).

    The plot visually demonstrates the relationship between a function and its antiderivative, illustrating how the accumulation of areas under the curve leads to the antiderivative function.


  2. Definite Integral:

    For a continuous function \( f \) on an interval \([a, b]\), the definite integral \( \int_a^b f(x) \, dx \) denotes the signed area enclosed by the graph of \( y = f(x) \), the x-axis, and the lines \( x = a \) and \( x = b \).

    By the fundamental theorem of calculus, we have \( \int_a^b f(x) \, dx = F(b) - F(a) \), where \( F \) is any antiderivative of \( f \).

    Note: In the expression \( \int_a^b f(x) \, dx \), the number \( a \) is called the lower limit of integration and \( b \) the upper limit of integration. The function \( f \) is called the integrand.

    Consider the function \( f(x) = x^2 \) defined on the interval \([0, 2]\). The plot visually represents this function.

    To illustrate the definite integral of \( f(x) \) over the interval \([0, 2]\), we shade the region under the curve \( f(x) \) between \( x = 0 \) and \( x = 2 \). This shaded area represents the signed area enclosed by the graph of \( f(x) \), the x-axis, and the vertical lines \( x = 0 \) and \( x = 2 \).

    The shaded region visually demonstrates the concept of a definite integral, which calculates the signed area between a function and the x-axis over a specific interval.


    Integration Formulas

    Type of Function Formula
    Sum Rule \( \int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx \)
    Difference Rule \( \int [f(x) - g(x)] \, dx = \int f(x) \, dx - \int g(x) \, dx \)
    Constant \( \int k \, dx = kx + C \)
    Power Functions \( \int x^n \, dx = \frac{1}{n+1} x^{n+1} + C \)
    Exponential \( \int e^x \, dx = e^x + C \)
    Logarithmic \( \int \ln(x) \, dx = x \ln(x) - x + C \)
    Linear Function \( \int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln |ax + b| + c \) for \( ax + b \neq 0 \)
    Trigonometric \( \int \sin(x) \, dx \) = \( -\cos(x) + C \)
    \( \int \cos(x) \, dx \) = \( \sin(x) + C \)
    \( \int \tan(x) \, dx \) = \( -\ln|\cos(x)| + C \)
    \( \int \cot(x) \, dx \) = \( \ln|\sin(x)| + C \)
    \( \int \sec(x) \, dx \) = \( \ln|\sec(x) + \tan(x)| + C \)
    \( \int \csc(x) \, dx \) = \( -\ln|\csc(x) + \cot(x)| + C \)
    Inverse Trigonometric \( \int \ sin^{-1}\left (x \right) \, dx \) = \( x\sin^{-1}\left (x \right) + \sqrt{1 - x^2} + C \)
    \( \int \cos^{-1}\left (x \right) \, dx \) = \( x\ cos^{-1}\left (x \right) - \sqrt{1 - x^2} + C \)
    \( \int \tan^{-1}\left (x \right) \, dx \) = \( x\text tan^{-1}\left (x \right) - \frac{1}{2}\ln(1+x^2) + C \)
    \( \int \text sec^{-1}\left (x \right) \, dx \) = \( x\text sec^{-1}\left (x \right) - \ln|x+\sqrt{x^2-1}| + C \)
    \( \int \text csc^{-1}\left (x \right) \, dx \) = \( x\text csc^{-1}\left (x \right) + \ln|x+\sqrt{x^2-1}| + C \)
    \( \int \text cot^{-1}\left (x \right) \, dx \) = \( x\text cot^{-1}\left (x \right) + \frac{1}{2}\ln(1+x^2) + C \)
    Hyperbolic \( \int \text sinh^{-1}\left (x \right) \, dx \) = \( x\text sinh^{-1}\left (x \right) - \sqrt{x^2+1} + C \)
    \( \int \text cosh^{-1}\left (x \right) \, dx \) = \( x\text cosh^{-1}\left (x \right) - \sqrt{x^2-1} + C \)
    \( \int \text tanh^{-1}\left (x \right) \, dx \) = \( x\text tanh^{-1}\left (x \right) + \frac{1}{2}\ln(1-x^2) + C \)
    \( \int \text sech^{-1}\left (x \right) \, dx \) = \( x\text sech^{-1}\left (x \right) - \ln|x+\sqrt{x^2-1}| + C \)
    \( \int \text sch^{-1}\left (x \right) \, dx \) = \( x\text csch^{-1}\left (x \right) + \ln|x+\sqrt{x^2+1}| + C \)
    \( \int \text coth^{-1}\left (x \right) \, dx \) = \( x\text coth^{-1}\left (x \right) + \frac{1}{2}\ln(x^2-1) + C \)
    Error Function \( \int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \text{erf}(x) + C \)
    Definite Integrals \( \int_a^b f(x) \, dx = \text{Area under } f(x) \text{ between } x=a \text{ and } x=b \)
    Indefinite Integrals \[ \int f(x) \, dx = F(x) + C \]

Graphs of functions and their antiderivative

In each of the examples in this section, the functions \( F \) and \( f \) are such that \( F'(x) = f(x) \). That is, the function \( F \) is an antiderivative of \( f \).

Consider the graphs of \( y = f(x) \) and \( y = F(x) \) shown. Find:

  1. a) f(x)
  2. b) F(x)

Solution

f(x) = mx

Since \( f(1) = 1 \), we have \( m = 1 \).

Hence \( f(x) = x \).

  1. F(x) = \( \frac{{x^2}}{2} + c \) (by antidifferentiation)
  2. But \( F(0) = 1 \) and therefore \( c = 1 \).

Hence \( F(x) = \frac{{x^2}}{2} + 1 \).

The graph of \( y = f(x) \) is the gradient graph for the graph of \( y = F(x) \).

Applications of Integration

Applications of integration are vast and diverse, spanning across various fields of science, engineering, economics, and more. Some common applications of integration include:

  • Area and Volume Calculations: Integration is used to find areas under curves and volumes of irregular shapes.
  • Physics: Integration helps calculate work, energy, and fluid flow in physical systems.
  • Engineering: Integration is essential for stress analysis, heat transfer, and designing electrical circuits.
  • Economics: Integration is applied to calculate consumer and producer surplus and optimize business decisions.
  • Probability and Statistics: Integration is used to analyze random variables, probability distributions, and solve statistical problems.
  • Signal Processing: Integration aids in analyzing signals, such as decomposing them into frequency components.
  • Biological and Medical Sciences: Integration helps model biological processes and analyze medical data, such as EEG signals.
  • Finance: Integration assists in calculating present and future values of cash flows and determining option pricing models.
Example 1

Find the integration of \(6x^3 - \frac{2}{x^2}\) with respect to \(x\).

To find the integration of \(6x^3 - \frac{2}{x^2}\) with respect to \(x\), you can integrate each term separately:

  1. For \(\int 6x^3 \, dx\), use the power rule for integration:
    \(\int 6x^3 \, dx = \frac{6}{4}x^4 + C_1 = \frac{3}{2}x^4 + C_1\)
  2. For \(\int \frac{2}{x^2} \, dx\), rewrite \(\frac{2}{x^2}\) as \(2x^{-2}\) and apply the power rule for integration:
    \(\int \frac{2}{x^2} \, dx = 2 \int x^{-2} \, dx = 2 \left( \frac{x^{-2+1}}{-2+1} \right) + C_2 = -2x^{-1} + C_2 = -\frac{2}{x} + C_2\)

Therefore, the integral of \(6x^3 - \frac{2}{x^2}\) with respect to \(x\) is:
\(\frac{3}{2}x^4 - \frac{2}{x} + C\)
where \(C\) is the constant of integration.

Example 2

Determine the indefinite integral of \( e^x - \sin(x) \).

To find the indefinite integral of \( e^x - \sin(x) \) with respect to \( x \), we integrate each term separately:

  1. Integral of \( e^x \): \[ \int e^x \, dx = e^x + C_1 \]
  2. Integral of \( -\sin(x) \): \[ \int -\sin(x) \, dx = \cos(x) + C_2 \]

Putting them together, the indefinite integral is: \[ \int (e^x - \sin(x)) \, dx = e^x - \cos(x) + C \]

where \( C = C_1 + C_2 \) is the constant of integration.

Example 3

Find the antiderivative w.r.t x \( \int e^{3x+1} \, dx \).

We have the integral \( \int e^{3x+1} \, dx \).

Let \( u = 3x + 1 \), then \( du = 3 \, dx \).

Solving for \( dx \), we get \( dx = \frac{1}{3} \, du \).

Now we can rewrite the integral in terms of \( u \):

\[ \int e^u \cdot \frac{1}{3} \, du \]

This simplifies to:

\[ \frac{1}{3} \int e^u \, du = \frac{1}{3} e^u + C \]

Now, substituting back \( u = 3x + 1 \), we get:

\[ \frac{1}{3} e^{3x + 1} + C \]

So, the solution to the integral \( e^{3x+1} \, dx \) is \( \frac{1}{3} e^{3x + 1} + C \).

Example 4

Evaluate the following integral: \( \int_0^1 (e^{2x} - e^{x}) \, dx \).

\[ \int_{0}^{1} (e^{2x} - e^{x}) \, dx = \left[\frac{1}{2} e^{2x} - e^{x}\right]_{0}^{1} \\ = \left[\frac{1}{2} e^{2} - e - \left(\frac{1}{2} - 1\right)\right] \\ = \frac{e^2}{2} - e + \frac{1}{2} \]

Example 5

Evaluate the following integral: \( \int_0^{\frac{\pi}{8}} \sec^2(2x) \, dx \).



We know that if \( f(x) = \tan(ax + b) \), then \( f'(x) = a \sec^2(ax + b) \). Hence \( \int \sec^2(ax + b) \, dx = \frac{1}{a} \tan(ax + b) + c \).

Therefore, \( \int_{0}^{\frac{\pi}{8}} \sec^2(2x) \, dx = \left[ \frac{1}{2} \tan(2x) \right]_{0}^{\frac{\pi}{8}} = \frac{1}{2} \left( \tan\left(\frac{\pi}{4}\right) - \tan(0) \right) = \frac{1}{2} (1 - 0) = \frac{1}{2} \).

Example 6

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

\[ \int_{-2}^{-1} (4x + 2) \, dx = \left( \frac{1}{4} \log_e |4x + 2| \right) \Bigg|_{-2}^{-1} = \frac{1}{4} \left( \log_e |{-2}| - \log_e |{-6}| \right) = \frac{1}{4} \log_e \left( \frac{2}{6} \right)=\frac{1}{4} \log_e \left( \frac{1}{3} \right) = -\frac{1}{4} \log_e 3 \]

Example 7

Evaluate the indefinite integral:\[ \int \frac{1}{9 + 16t^2} \, dt \]

\[ \int \frac{1}{9 + 16t^2} \, dt \] \[ = \int \frac {1}{16(\frac{9}{16}+t^2) }\, dt \]
\[ = \frac{4}{3} \int \frac{\frac{3}{4}}{16(\frac{9}{16} + t^2)} \, dt \]
\[ = \frac{1}{2} \int \frac{\frac{3}{4}}{\frac{9}{16} + t^2} \, dt \]
\[ = \frac{1}{2} \tan^{-1} \left( \frac{4t}{3} \right) +C\]

Example 8
Evaluate the definite integral:\[ \int_{0}^{1} \frac{3}{\sqrt{9 - 4x^2}} \, dx \]
\[ = \int_{0}^{1} \frac {3}{2\sqrt{\frac{9}{4} - x^2}} \, dx \] \[ = \frac{3}{2} \int_{0}^{1} \sqrt{\frac{9}{4} - x^2} \, dx \] \[ = \frac{3}{2} \left[ \sin^{-1} \left( \frac{2x}{3} \right) \right]_{0}^{1} \] \[ = \frac{3}{2} \sin^{-1} \left( \frac{2}{3} \right) \] ≈ 1.095
Example 9

The graph of \( y = f(x) \) is as shown.

Sketch the graph of \( y = F(x) \), given that \( F(0) = 0 \).

The given graph \( y = f(x) \) is the gradient graph of \( y = F(x) \).

Therefore, the gradient of \( y = F(x) \) is always positive.

The minimum gradient is 2 and this occurs when \( x = -1 \).

There is a line of symmetry \( x = -1 \), which indicates equal gradients for \( x \)-values equidistant from \( x = -1 \).

Also, \( F(0) = 0 \).

A possible graph is shown.

Example 10

Find the integral of \( \frac{1}{\cosh^2(x)} \, dx \).

We'll start by using the identity \( \cosh^2(x) - \sinh^2(x) = 1 \), so \( \cosh^2(x) = 1 + \sinh^2(x) \).

Now, let's substitute \( \sinh(x) = t \), which implies \( \cosh(x) \, dx = dt \).

The integral becomes: \[ \int \frac{1}{\cosh^2(x)} \, dx = \int \frac{1}{1 + t^2} \, dt \]

This integral can be evaluated using the tan^{-1} function: \[ = \tan^{-1}(t) + C \]

Substituting back \( t = \sinh(x) \), we get: \[ = \tan^{-1}(\sinh(x)) + C \]

So, the integral of \( \frac{1}{\cosh^2(x)} \) with respect to \( x \) is \( \tan^{-1}(\sinh(x)) + C \).

Exercise &&1&& (&&1&& Question)

Find the indefinite integral of \(3x^2 - 2x + 5\).

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Exercise &&2&& (&&1&& Question)

Evaluate the definite integral \( \int_{0}^{1} (2x + 3) \, dx \).

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Exercise &&3&& (&&1&& Question)

Compute the area enclosed by the curve \( y = e^x \) and the x-axis from \( x = 0 \) to \( x = 2 \).

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Exercise &&4&& (&&1&& Question)

Find an antiderivative of the following: \( \sin(3x - \frac{\pi}{4}) \).

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Exercise &&5&& (&&1&& Question)

Evaluate the definite integral: \( \int_{0}^{2} \frac{1}{(4 + x^2)} \, dx \).

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Exercise &&6&& (&&1&& Question)

Compute the area enclosed by the curve \( y = x^3 \) and the x-axis from \( x = 0 \) to \( x = 2 \).

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Exercise &&7&& (&&1&& Question)

Evaluate the definite integral

\[ \int_{1}^{e} \frac{1}{x} \, dx. \]

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Exercise &&8&& (&&1&& Question)

Calculate the work done by a force \( F(x) = 3x^2 \) moving an object from \( x = 1 \) to \( x = 4 \).

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