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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: \[\text{if} \quad f(x) = a.x^n\] \[\begin{aligned} \text{then} \quad F(x) &= \int a.x^n dx \\ F(x) &= \frac{a}{n+1}x^{n+1} + c \end{aligned}\]

Tutorial 1

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

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