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# Reverse power rule review

Review your knowledge of the reverse power rule for integrals and solve problems with it.

## What is the reverse power rule?

The reverse power rule tells us how to integrate expressions of the form x, start superscript, n, end superscript where n, does not equal, minus, 1:
integral, x, start superscript, n, end superscript, d, x, equals, start fraction, x, start superscript, n, plus, 1, end superscript, divided by, n, plus, 1, end fraction, plus, C
Basically, you increase the power by one and then divide by the power plus, 1.
Remember that this rule doesn't apply for n, equals, minus, 1.
Instead of memorizing the reverse power rule, it's useful to remember that it can be quickly derived from the power rule for derivatives.

## Integrating polynomials

We can use the reverse power rule to integrate any polynomial. Consider, for example, the integration of the monomial 3, x, start superscript, 7, end superscript:
\begin{aligned} \displaystyle\int 3x^7\,dx&=3\left(\dfrac{x^{7+1}}{7+1}\right)+C \\\\ &=3\left(\dfrac{x^8}{8}\right)+C \\\\ &=\dfrac{3}{8}x^8+C \end{aligned}
Problem 1
integral, 14, t, d, t, equals, question mark

Want to try more problems like this? Check out these exercises:

## Integrating negative powers

The reverse power rule allows us to integrate any negative power other than minus, 1. Consider, for example, the integration of start fraction, 1, divided by, x, squared, end fraction:
\begin{aligned} \displaystyle\int \dfrac{1}{x^2}\,dx&=\displaystyle\int x^{-2}\,dx \\\\ &=\dfrac{x^{-2+1}}{-2+1}+C \\\\ &=\dfrac{x^{-1}}{-1}+C \\\\ &=-\dfrac{1}{x}+C \end{aligned}
Problem 1
integral, 8, t, start superscript, minus, 3, end superscript, d, t, equals

Want to try more problems like this? Check out these exercises:

## Integrating fractional powers and radicals

The reverse power rule also allows us to integrate expressions where x is raised to a fractional power, or radicals. Consider, for example, the integration of square root of, x, end square root:
\begin{aligned} \displaystyle\int \sqrt x\,dx&=\displaystyle\int x^{^{\large\frac{1}{2}}}\,dx \\\\ &=\dfrac{x^{^{\large\frac{1}{2}\normalsize+1}}}{\dfrac{1}{2}+1}+C \\\\ &=\dfrac{x^{^{\large\frac{3}{2}}}}{\frac{3}{2}}+C \\\\ &=\dfrac{2\sqrt{x^3}}{3}+C \end{aligned}
Problem 1
integral, 4, t, start superscript, start fraction, 1, divided by, 3, end fraction, end superscript, d, t, equals, question mark