1.8.1 Basics of Integration - GCEALevelMaths9709

In this topic we will learn how to:

understand integration as the reverse process of differentiation, and integrate $(ax + b)^{n}$ (for any rational $n$ except $-1$ , together with constant multiples, sums and differences
solve problems involving the evaluation of the constant of integration, including to the find the equation of a curve

Integration helps us sum up areas, and volumes under curves. In simple terms, it helps us find areas and volumes for irregular shapes, defined by functions. Integration is the reverse process of differentiation. The notation for integration is displayed below,
$\int f(x)\ dx$ The above reads as ‘The integral of $f(x)$ with respect to $x$ ‘. This is known as an indefinite integral, since we have not defined any limits. Let’s define some limits,
$\int_{a}^{b} f(x)\ dx$ The above reads as ‘The integral of $f(x)$ , from $a$ to $b$ with respect to $x$ ‘. This is known as a definite integral, since we have defined some limits. Definite integrals are used to evaluate the area under a graph from point $a$ to point $b$ . $\textcolor{gray}{\textbf{Integration of }x^{n}}$ To integrate $x^{n}$ we use the general formula,
$\frac{x^{n + 1}}{n + 1}$ Let’s look at an example.

Example 1
Find $\int x^{3}\ dx$ .

Add $1$ to the power and divide by the new power,
$\frac{x^{3 + 1}}{3 + 1}$ $\frac{x^{4}}{4}$

$\textcolor{gray}{\textbf{Integration of }(ax + b)^{n}}$ To integrate $(ax + b)^{n}$ we use the general formula,
$\frac{(ax + b)^{n + 1}}{a(n + 1)}$ Let’s look at an example.

Example 2
Find $\int (2x + 3)^{4}$ .
Add $1$ to the power and divide by the new power,
$\frac{(2x + 3)^{4 + 1}}{2(4 + 1)}$ $\frac{(2x + 3)^{5}}{2(5)}$ $\frac{(2x + 3)^{5}}{10}$

Let’s look at some past paper questions.

1. The equation of a curve is such that $\frac{dy}{dx} = 3(4x - 7)^{\frac{1}{2}} - 4x^-{\frac{1}{2}}$ . It is given that the curve passes through the point $\left(4, \frac{5}{2}\right)$ . Find the equation of the curve. (9709/12/M/J/22 number 3)

To find the equation of the curve we have to integrate $\frac{dy}{dx}$ ,
$\int \frac{dy}{dx}\ dx= 3(4x - 7)^{\frac{1}{2}} dx - 4x^{-\frac{1}{2}}\ dx$ Note: This is an indefinite integral, as no limits are defined.
$y = \frac{3(4x - 7)^{\frac{3}{2}}}{4\left(\frac{3}{2}\right)} - \frac{4x^{\frac{1}{2}}}{\frac{1}{2}} + c$ Note: $c$ represents the constant of integration. Whenever we evaluate an indefinite integral we add the constant of integration, which we also have to evaluate.
$y = \frac{1}{2}(4x - 7)^{\frac{3}{2}} - 8x^{\frac{1}{2}} + c$ To evaluate $c$ let’s substitute the coordinates of point on the curve, $\left(4, \frac{5}{2}\right)$ ,
$y = \frac{1}{2}(4x - 7)^{\frac{3}{2}} - 8x^{\frac{1}{2}} + c$ $\frac{5}{2} = \frac{1}{2}(4(4) - 7)^{\frac{3}{2}} - 8(4)^{\frac{1}{2}} + c$ $\frac{5}{2} = -\frac{5}{2} + c$ $c = \frac{5}{2} + \frac{5}{2}$ $c = 5$ Therefore, the equation of the curve is,
$y = \frac{1}{2}(4x - 7)^{\frac{3}{2}} - 8x^{\frac{1}{2}} + 5$
2. Find $\int_{0}^{1}\ x^{-\frac{1}{2}}\ dx$ .
$\int_{0}^{1}\ x^{-\frac{1}{2}}\ dx$ This is a definite integral since the limits are defined. Let’s integrate it,
$\int_{0}^{1}\ x^{-\frac{1}{2}}\ dx$ $\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]{0}^{1}$ $\left[2x^{\frac{1}{2}}\right]{0}^{1}$ Substitute in the limits and simplify,
$\left[2(1)^{\frac{1}{2}} - 2(0)^{\frac{1}{2}}\right]$ $2$ Note: For a definite integral we do not put the constant of integration, because it would cancel out.

Therefore, the final answer is,
$\int_{0}^{1}\ x^{-\frac{1}{2}}\ dx = 2$ 3. Find $\int_{1}^{\infty}\ (4x + 2)^{-2}\ dx$ . (9709/13/M/J/22 number 2)

This is an indefinite integral. Let’s integrate,
$\int_{1}^{\infty}\ (4x + 2)^{-2}\ dx$ $\left[\frac{(4x + 2)^{-1}}{(4)(-1)}\right]_{1}^{\infty}$ $\left[\frac{-1}{4(4x + 2)}\right]_{1}^{\infty}$ $\left[\frac{-1}{16x + 8}\right]_{1}^{\infty}$ Substitute in the limits,
$\left[\left(\frac{-1}{16(\infty) + 8}\right) - \left(\frac{-1}{16(1) + 8}\right)\right]$ $\left[0 - \left(\frac{-1}{24}\right)\right]$ $\frac{1}{24}$ Note: $\infty$ represents a very large number such that $\frac{1}{\infty}$ is assigned a value of $0$ . If you input $1$ divided by a very big number into a calculator, you will get a very small number, close to $0$ , that’s why $\frac{1}{\infty} = 0$ .

Therefore, the final answer is,
$\int_{1}^{\infty}\ (4x + 2)^{-2}\ dx = \frac{1}{24}$

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