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# Power Rule for Integration

The power rule for integration provides us with a formula that allows us to integrate any function that can be written as a power of $$x$$.
By the end of this section we'll know how to evaluate integrals like: $\int 4x^3 dx$ $\int \frac{3}{x^2}dx$ $\int \begin{pmatrix} 2x + 3 \sqrt{x} \end{pmatrix} dx$ We start by learning the power rule for integration formula, before watching a tutorial and working through some exercises.
We also treat each of the "special cases" such as negatitive and fractional exponents to integrate functions involving roots and reciprocal powers of $$x$$.
The power rule for integration is an essential step in learning integration, make sure to work through all of the exercises and to watch all of the tutorials.

## Power Rule for Integration

Given a function, which can be written as a power of $$x$$, we can integrate it using the power rule for integration: $\text{if} \quad f(x) = a.x^n$ \begin{aligned} \text{then} \quad F(x) &= \int a.x^n dx \\ F(x) &= \frac{a}{n+1}x^{n+1} + c \end{aligned}

## Exercise 1

Integrate each of the following

1. $$\int 5x^3 dx$$

2. $$\int 3x dx$$

3. $$\int 7x^2 dx$$

4. $$\int -4x^5 dx$$

5. $$\int 12 x^7 dx$$

6. $$\int -2x^5 dx$$

7. $$\int 2 dx$$

8. $$\int 10 x^4 dx$$

## Solution

1. $$\int 5x^3 dx = \frac{5}{4}x^4 + c$$

2. $$\int 3x dx = \frac{3}{2}x^2 + c$$

3. $$\int 7x^2 dx = \frac{7}{3}x^3 + c$$

4. $$\int -4x^5 dx = -\frac{2}{3}x^6 + c$$

5. $$\int 12x^7 dx = \frac{3}{2}x^8 + c$$

6. $$\int -2x^5 dx = -\frac{1}{3}x^6 + c$$

7. $$\int 2 dx = 2x + c$$

8. $$\int 10x^4 dx = 2x^5 + c$$

## Solution

1. We integrate $$5x^3$$ as follows: \begin{aligned} \int 5x^3 dx &= \frac{5}{3+1}x^{3+1}dx \\ & = \frac{5}{4}x^4 + c \end{aligned}

2. We integrate $$3x$$ using the power rule for integration and the fact that $$x = x^1$$: \begin{aligned} \int 3x dx & = \int 3x^1 dx \\ & = \frac{3}{1+1}x^{1+1} + c \\ & = \frac{3}{2}x^2 + c \end{aligned}

3. We integrate $$7x^2$$ as follows: \begin{aligned} \int 7x^2 dx & = \frac{7}{2+1}x^{2+1} + c \\ & = \frac{7}{3}x^3 + c \end{aligned}

4. We integrate $$-4x^5$$ as follows: \begin{aligned} \int - 4x^5 dx & = -\frac{4}{5+1}x^{5+1} +c \\ & = - \frac{4}{6}x^6 + c \\ & = - \frac{2}{3}x^6 + c \end{aligned}

5. We integrate $$12x^7$$ as follows: \begin{aligned} \int 12x^7 dx & = \frac{12}{7+1}x^{7+1} + c \\ & = \frac{12}{8}x^8 + c \\ & = \frac{3}{2}x^8 + c \end{aligned}

6. We integrate $$-2x^5$$ as follows: \begin{aligned} \int -2x^5 dx & = - \frac{2}{5+1}x^{5+1} + c \\ & = - \frac{2}{6}x^{6} + c \\ & = - \frac{1}{3} x^6 + c \end{aligned}

7. We integrate $$2$$ using the fact that $$2 = 2x^0$$: \begin{aligned} \int 2 dx & = \int 2x^0 dx \\ & = \frac{2}{0+1}x^{0+1} + c \\ & = \frac{2}{1}x^1+c \\ & = 2x+c \end{aligned}

8. We integrate $$10x^4$$ as follows: \begin{aligned} \int 10x^4 dx & = \frac{10}{4+1}x^{4+1} + c \\ & = \frac{10}{5}x^5+c\\ & = 2x^5+c \end{aligned}

## Negative Exponents

Any function looking like $$f(x) = \frac{a}{x^n}$$ can be written using a negative exponent: $f(x) = a.x^{-n}$ Using this fact we can integrate any function written as: $f(x) = \frac{a}{x^n}$ Except for $$\frac{a}{x}$$! Indeed, we'll soon learn about that special case.

This formula is illustrated wih some worked examples in Tutorial 2.

Now that we've seen that we can integrate functions looking like $$f(x)=\frac{a}{x^n}$$ using negative powers of $$x$$, let's work through the exercise below.

Useful Trick: it's often useful to use the fact that $$\int ax^n dx = a \int x^n dx$$, particularly when $$a$$ is a fraction like in question 5. in the following exercise. It's useful to write: $$\int \frac{5}{2x^3}dx = \frac{5}{2}\int \frac{1}{x^3}dx$$ to not let the fraction $$\frac{5}{2}$$ lead to a error in arithmetic.

## Exercise 2

1. $$\int \frac{2}{x^3}dx$$

2. $$\int \frac{3}{x^5}dx$$

3. $$\int -\frac{1}{x^2} dx$$

4. $$\int \frac{6}{x^5}dx$$

5. $$\int \frac{5}{2x^3} dx$$

## Solution Without Working

1. $$\int \frac{2}{x^3} dx = - \frac{1}{x^2} + c$$

2. $$\int \frac{3}{x^5} dx = - \frac{3}{4x^4} + c$$

3. $$\int - \frac{1}{x^2} dx = \frac{1}{x} + c$$

4. $$\int \frac{6}{x^5} dx = - \frac{3}{2x^4} + c$$

5. $$\int \frac{5}{2x^3} dx = \frac{2}{7x^2} + c$$

## Solutions With Working

1. To integrate $$\frac{2}{x^3}$$ we use the fact that $$\frac{2}{x^3} = 2x^{-3}$$: \begin{aligned} \int \frac{2}{x^3} dx &= \int 2x^{-3} dx \\ & = \frac{2}{-3+1}x^{-3+1}+c \\ & = \frac{2}{-2}x^{-2}+c \\ & = -x^{-2}+c \\ & = - \frac{1}{x^2} + c \end{aligned}

2. To integrate $$\frac{3}{x^5}$$ we use the fact that $$\frac{3}{x^5} = 3x^{-5}$$: \begin{aligned} \int \frac{3}{x^5} dx & = \int 3x^{-5} dx \\ & = \frac{3}{-5+1}x^{-5+1} + c \\ & = \frac{3}{-4}x^{-4} + c \\ & = - \frac{3}{4} x^{-4} + c \\ & = - \frac{3}{4}.\frac{1}{x^4} + c \\ & = - \frac{3}{4x^4} + c \end{aligned}

3. To integrate $$-\frac{1}{x^2}$$ we use the fact that $$-\frac{1}{x^2} = -x^{-2}$$: \begin{aligned} \int - \frac{1}{x^2} dx & = \int -x^{-2}dx \\ & = \frac{-1}{-2+1}x^{-2+1} + c \\ & = \frac{-1}{-1}x^{-1} +c \\ & = x^{-1} + c \\ & = \frac{1}{x} + c \end{aligned}

4. To integrate $$\frac{6}{x^5}$$ we use the fact that $$\frac{6}{x^5} = 6x^{-5}$$: \begin{aligned} \int \frac{6}{x^5} dx & = \int 6x^{-5} dx \\ & = \frac{6}{-5+1}x^{-5+1} + c \\ & = \frac{6}{-4}x^{-4} + c \\ & = - \frac{3}{2}x^{-4}+ c \\ & = - \frac{3}{2}.\frac{1}{x^4} + c \\ & = - \frac{3}{2x^4} + c \end{aligned}

5. To integrate $$\frac{5}{2x^3}$$, we use the fact that $$\frac{5}{2x^3} = \frac{5}{2}x^{-3}$$: \begin{aligned} \int \frac{5}{2x^3} dx & = \int \frac{5}{2}x^{-3} dx \\ & = \frac{5}{2}\int x^{-3}dx \\ & = \frac{5}{2} \times \frac{1}{-3+1}x^{-3+1}+c \\ & = \frac{5}{2} \times \frac{1}{-2}x^{-2}+c \\ & = \frac{5}{-4}.x^{-2}+c \\ & = - \frac{5}{4}.x^{-2}+c\\ & = - \frac{5}{4}.\frac{1}{x^2}+c \\ \int \frac{5}{2x^3} dx & = - \frac{5}{4x^2}+c \end{aligned}

## Fractional Exponents

Functions looking like $$f(x) = a.\sqrt[n]{x^m}$$ can be written as powers of $$x$$ using fractional exponents: $f(x) = a.x^{\frac{m}{n}}$ we can therefore use the power rule for integration to integrate any function looking like $$f(x)=a.\sqrt[n]{x^m}$$.

This formula is illustrated wih some worked examples in Tutorial 3.

Now that we've seen how to integrate roots using fractional powers of $$x$$, let's work through a few more questions.

## Exercise 3

Integrate each of the following:

1. $$\int \sqrt{x} dx$$

2. $$\int 2.\sqrt[3]{x} dx$$

3. $$\int 4. \sqrt[5]{x^4}dx$$

4. $$\int 6 .\sqrt{x}dx$$

5. $$\int \frac{\sqrt[3]{x}}{4} dx$$

## Solutions Without Working

1. The answer can be written in two ways:

$$\int \sqrt{x} dx = \frac{2}{3} \sqrt{x^3}+c$$ or $$\int \sqrt{x} dx = \frac{2}{3}.x^{\frac{3}{2}}+c$$

2. The answer can be written in two ways:

$$\int 2. \sqrt[3]{x} dx = \frac{3}{2} \sqrt[3]{x^4}+c$$ or $$\int 2. \sqrt[3]{x} dx = \frac{3}{2}.x^{\frac{3}{4}}+c$$

3. The answer can be written in two ways:

$$\int 4.\sqrt[5]{x^4}dx = \frac{20}{9}.\sqrt[5]{x^9}+c$$ or $$\int 4.\sqrt[5]{x^4}dx = \frac{20}{9}.x^{\frac{9}{5}}+c$$

4. The answer can be written in two ways:

$$\int 6 \sqrt{x}dx = 4.\sqrt{x^3}+c$$ or $$\int 6 \sqrt{x}dx = 4.x^{\frac{3}{2}}+c$$

5. The answer can be written in two ways:

$$\int \frac{\sqrt[3]{x}}{4}dx = \frac{3}{16}.\sqrt[3]{x^4}+c$$ or $$\int \frac{\sqrt[3]{x}}{4}dx = \frac{3}{16}.x^{\frac{4}{3}}+c$$

## Solutions With Working

1. We integrate $$\int \sqrt{x} dx$$ as follows: \begin{aligned} \int \sqrt{x} dx & = \int x^{\frac{1}{2}} dx \\ & = \frac{1}{\frac{1}{2} + 1}.x^{\frac{1}{2}+1} + c \\ & = \frac{1}{\frac{3}{2}}.x^{\frac{3}{2}}+c \\ & = \frac{2}{3}.x^{\frac{3}{2}}+c \\ \int \sqrt{x} dx & = \frac{2}{3}.\sqrt{x^3}+c \end{aligned}

2. We integrate $$\int 2. \sqrt[3]{x} dx$$ as follows: \begin{aligned} \int 2. \sqrt[3]{x} dx & = \int 2.x^{\frac{1}{3}} dx \\ & = \frac{2}{\frac{1}{3}+1}.x^{\frac{1}{3}+1}+c \\ & = \frac{2}{\frac{4}{3}}x^{\frac{4}{3}}+c \\ & = 2 \times \frac{3}{4}.x^{\frac{4}{3}} + c \\ & = \frac{3}{2}.x^{\frac{4}{3}}+c \\ \int 2. \sqrt[3]{x} dx & = \frac{3}{2}.\sqrt[3]{x^4}+c \end{aligned}

3. We integrate $$\int 4 \sqrt[5]{x^4} dx$$ as follows: \begin{aligned} \int 4 \sqrt[5]{x^4} dx & = \int 4.x^{\frac{4}{5}} dx \\ & = \frac{4}{\frac{4}{5}+1}.x^{\frac{4}{5}+1}+c \\ & = \frac{4}{\frac{9}{5}}.x^{\frac{9}{5}}+c \\ & = 4 \times \frac{5}{9}.x^{\frac{9}{5}}+c \\ & = \frac{20}{9}.x^{\frac{9}{5}}+C \\ \int 4 \sqrt[5]{x^4} dx & = \frac{20}{9}.\sqrt[5]{x^9} + c \end{aligned}

4. We integrate $$\int 6.\sqrt{x} dx$$ as follows: \begin{aligned} \int 6.\sqrt{x} dx & = \int 6.x^{\frac{1}{2}}+c \\ & = 6\times \frac{1}{\frac{1}{2}+1}.x^{\frac{1}{2}+1}+c \\ & = 6 \times \frac{1}{\frac{3}{2}}.x^{\frac{3}{2}}+c \\ & = 6 \times \frac{2}{3}.x^{\frac{3}{2}}+c \\ & = 4.x^{\frac{3}{2}}+c \\ \int 6.\sqrt{x} dx & = 4.\sqrt{x^3}+c \end{aligned}

5. We integrate $$\int \frac{\sqrt[3]{x}}{4} dx$$ as follows: \begin{aligned} \int \frac{\sqrt[3]{x}}{4} dx & = \frac{1}{4} \int \sqrt[3]{x} dx \\ & = \frac{1}{4} \int x^{\frac{1}{3}} dx \\ & = \frac{1}{4} \times \frac{1}{\frac{1}{3}+1}.x^{\frac{1}{3}+1}+c \\ & = \frac{1}{4} \times \frac{1}{\frac{4}{3}}x^{\frac{4}{3}}+c \\ & = \frac{1}{4}\times \frac{3}{4}.x^{\frac{4}{3}}+c \\ & = \frac{3}{16}.x^{\frac{4}{3}}+c\\ \int \frac{\sqrt[3]{x}}{4} dx & = \frac{3}{16}.\sqrt[3]{x^4}+c \end{aligned}

### Integrands with more than one term

We now look at integrals in which the integrand has more than one term.
For instance: $$\int \begin{pmatrix} x^2 + x^3 \end{pmatrix}dx$$.
To evaluate such integrals, we integrate each term as though it was on its own: $\int \begin{pmatrix} x^2 + x^3 \end{pmatrix}dx = \int x^2 dx + \int x^3 dx$ Keeping that in mind, let's work through the following exercise.

## Exercise 4

Integrate each of the following:

1. $$\int \begin{pmatrix} 4 - \frac{1}{x^2} \end{pmatrix}dx$$

2. $$\int \begin{pmatrix} x^2 + \frac{3}{x^3} \end{pmatrix}dx$$

3. $$\int \frac{4+2x}{x^3} dx$$

4. $$\int \frac{3x^3 - 1}{x^2} dx$$

5. $$\int \frac{dx}{4\sqrt{x}}$$

## Solution Without Working

1. We can write the solution in two ways:

$$\int \begin{pmatrix} 4 - \frac{1}{x^2} \end{pmatrix}dx = 4x + \frac{1}{x} + c$$, or

$$\int \begin{pmatrix} 4 - \frac{1}{x^2} \end{pmatrix}dx = 4x + x^{-1}+c$$

2. We can write the solution in two ways:

$$\int \begin{pmatrix} x^2 + \frac{3}{x^3} \end{pmatrix} dx = \frac{x^3}{3} - \frac{3}{2x^2}+c$$, or

$$\int \begin{pmatrix} x^2 + \frac{3}{x^3} \end{pmatrix} dx = \frac{1}{3}x^3 - \frac{3}{2}.x^{-2}+c$$

3. We can write the solution in two ways:

$$\int \frac{4+2x}{x^3} dx = - \frac{2}{x^2} - \frac{2}{x} + c$$, or

$$\int \frac{4+2x}{x^3} dx = -2.x^{-2}-2.x^{-1}+c$$

4. We can write the solution in two ways:

$$\int \frac{3x^3 - 1}{x^2} dx = \frac{3}{2}x^2 + \frac{1}{x}+c$$, or

$$\int \frac{3x^3 - 1}{x^2} dx = \frac{3}{2}x^2 + x^{-1}+c$$

5. We can write the solution in two ways:

$$\int \frac{dx}{4\sqrt{x}} = \frac{1}{2}\sqrt{x} + c$$, or

$$\int \frac{dx}{4\sqrt{x}} = \frac{1}{2}x^{\frac{1}{2}}+c$$