# THE POWER RULE OF INTEGRATION

Integration is the inverse of differentiation.

It is the process of finding the function whose derivative is known.

Because integration is the inverse of differentiation, it is also known as anti-derivative.

One way of integrating is using the power rule.

The power of the rule of integration is easily achieved by adding one to the exponent and then dividing by the value of the added exponent.

That is,

$$\int x dx=\frac{x^{n+1}}{n+1}$$

The power rule of integration allows us to integrate any function that can be written as a power of x.

And, always remember to add C. This ensures that the function derive can account for all possible solutions of the integral.

Example 1

Evaluate $\int 8x^5 dx$

Solution:

By power rule,

$\frac{8x^{5+1}}{5+1}+C$

$\frac{8x^6}{6}+C=\frac{4x^6}{3}$

Example 2

Evaluate $\int 4x^2+3x dx$

Solution:

By power rule

$\frac{4x^{2+1}}{2+1}+\frac{3x^{1+1}}{1+1}+C$

$\frac{4x^{3}}{3}+\frac{3x^2}{2}+C$

Example 3

Integrate $4x^3+6x^2+2$

Solution:

$\int 4x^3+6x^2+2dx$

By power rule

$\frac{4x^{3+1}}{3+1}+\frac{6x^{2+1}}{2+1}+\frac{2x^{0+1}}{0+1}+C$

$\frac{4x^{4}}{4}+\frac{6x^{3}}{3}+\frac{2x^{1}}{1}+C$

$x^4+2x^3+2x+C$

Example 4

Integrate $(1-x)^2$

Solution:

Before integrating, let's expand the expression

$(1-x)^2=1(1-x)-x(1-x)$

$(1-x)^2=1-x-x+x^2=1-2x+x^2$

We can now integrate

$\int 1-2x+x^2 dx$

$\frac{1x^{0+1}}{0+1}-\frac{2x^{1+1}}{1+1}+\frac{x^{2+1}}{2+1}+C$

$\frac{1x^1}{1}-\frac{2x^{2}}{2}+\frac{x^{3}}{3}+C$

$x-x^2++\frac{x^{3}}{3}+C$

Example 5

Evaluate $\int \left(\frac{2t^3-3t}{4t}\right) dx$

Solution:

Before integrating, let's simply the expression further

$\frac{2t^3-3t}{4t}=\frac{2t^3}{4t}-\frac{3t}{4t}$

$\frac{t^3}{4}-\frac{3}{4}$

Let's now integrate

$\int \left ({t^3}{4}-\frac{3}{4}\right) + C$

By power rule

$\frac{\frac{t^{2+1}}{2}}{2+1}-\frac{\frac{3t^{0+1}}{4}}{0+1}+C$

$\frac{\frac{t^{3}}{2}}{3}-\frac{3t^{1}}{4}+C$

$\frac{t^3}{6}-\frac{3t}{4}+C$

Example 6

Evaluate $\int \left(\frac{(3x^2-2)^2}{x^2}\right) dx$

Solution:

Before we integrate, let first simplify.

$\left(\frac{(3x^2-2)^2}{x^2}\right)$

Expanding the numerator

$\frac{3x^2(3x^2-2)-2(3x^2-2)}{x^2}$

$\frac{9x^4-6x^2-6x^2+4}{x^2}$

$\frac{9x^4-12x^2+4}{x^2}$

Splitting the fraction

$\frac{9x^4}{x^2}-\frac{12x^2}{x^2}+\frac{4}{x^2}$

$9x^{4-2}-12x^{2-2}+4x^{-2}$

$9x^2-12+4x^{-2}$

Having simplify, we can now integral

$\int 9x^2-12+4x^{-2} dx$

Using the power rule

$\frac{9x^{2+1}}{3}-\frac{12x^{0+1}}{1}+\frac{4x^{-2+1}}{-1}+C$

$3x^3-12x-4x^{-1}+C$

If you like, you can express in fractional form

$3x^3-12x-\frac{4}{x}+C$

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