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# Reverse power rule review

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.C (LO)
,
FUN‑6.C.1 (EK)
,
FUN‑6.C.2 (EK)
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

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

## Want to join the conversation?

• What would you do if n=-1? What would the integral be of x^-1?
• actually, ln|x| to include negative x values as well.
• how do you integrate sin^2 x?
• I'm wondering how to work the second part of fractional powers and radicals....specifically the 4 sqrtx^8...how did it become x^2? and also the first part, did it become x^(5/2) because I raised the function to (1/2) and then just multiply straight across on the exponents? And if I may ask one more question, On problem #3 integrating negative powers....on the 2t/t^3...I know division means subtraction...but can you walk me through that part please. Thank you for all your help.
• How do we go from 1/3 x^3 to 4sqrtx^8?
• 4sqrtx^8 is rewritten as x^2, because (x^2)^4 = x^8
Therefore, the antiderivative of x^2 is:
x^(2+1) / (2+1) + C
x^3 / (3) + C
1/3 x^3 + C
• It's so easy.
Just like differential calculus, integral calculus has its own rules.
• how do you do this
• What is integral of √ax+b dx
(1 vote)
• It is ambiguous.....are both ax under the radical or just a?....Let's solve the first case which is the most laborious case....

2*5^(1/2)*x^(3/2)/3 + bx I hope it helps!