The Power Rule for Antiderivatives

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Calculus $\longrightarrow$
Understanding Integration $\longrightarrow$

Objectives
• Learn the Power Rule for polynomial antiderivatives
• Be able to specifically state why it behaves like an inverse to the Power Rule for derivatives
• Practice taking definite and indefinite integrals of polynomials
Lesson Description

We recently learned that integration involves finding antiderivatives, so we need to know how to find them for any given function type. Here we will start with polynomials, using a Power Rule for integration similar to the one that exists for derivatives.

Practice Problems

Practice problems and worksheet coming soon!

Integrating Power Terms

Any power function of the form $f(x) = x^n$, where $n$ is any real number, can be quickly integrated, similarly to how those types of terms had a simple rule for derivatives.
Power Rule - AntiderivativesFor any real number constant $n$:$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$
This perfectly complements the Power Rule for derivatives », and we can see that the two rules are inverses of one another, as we expect them to be based on our introductory thoughts on antiderivatives ».Specifically, if $f(x) = x^n$, then$$F(x) = \int f(x) \, dx = \frac{x^{n+1}}{n+1} + C$$and$$\frac{d}{dx} \, F(x) = \frac{d}{dx} \, \frac{x^{n+1}}{n+1}$$$$=(n+1) \cdot \frac{x^{n+1-1}}{n+1} = x^n$$This shows us that the power rule for integrals "undoes" the power rule for derivatives.

Example 1Integrate $\displaystyle \int x^4 \, dx$.$\blacktriangleright$ Using the power rule:$$\int x^4 \, dx = \frac{x^{4+1}}{4+1} + C$$$$= \frac{x^5}{5}$$

Warning!
The Power Rule does not work if $n$ is $-1$. If we tried to apply it, we would get$$\int x^{-1} \, dx = \frac{x^0}{0} + C$$which is not defined. The antiderivative of $x^{-1}$ is $\ln(x)$, because the derivative of $\ln(x)$ is $1/x$. We'll see this again as we continue to study integration.

Using the Power Rule

Roots and RadicalsAs with the power rule » for derivatives, it's important to train your brain to look at radicals as fraction exponents » when doing calculus.After you take two seconds to re-write the problem using fraction exponents, the power rule can be applied directly.

Example 2$$\int \sqrt[3]{x^5} \, dx$$$\blacktriangleright$ Let's re-write this as a fraction exponent, and then apply the power rule.$$\int \sqrt[3]{x^5} \, dx = \int x^{5/3} \, dx$$$$= \frac{x^{5/3+1}}{5/3 + 1} + C$$$$= \frac{x^{8/3}}{8/3} + C$$$$= \frac{3x^{8/3}}{8} + C$$

Pro Tip
When you apply the power rule to integrate fraction exponent terms, the denominator has a fraction in it. The coefficient in the final answer will always be the reciprocal of that fraction. In the previous example, we had $8/3$ in the denominator, but the answer's coefficient was $3/8$. Use this to help you save time and answer accurately.
A similar idea is true for reciprocals - write them as negative exponents and the power rule can be applied comfortably.

Example 3$$\int \frac{1}{x^3} \, dx$$$\blacktriangleright$ Let's re-write this problem using negative exponents before using the power rule.$$\int \frac{1}{x^3} \, dx = \int x^{-3} \, dx$$$$= \frac{x^{-2}}{-2} + C$$$$= -\frac{1}{2x^2} + C$$

ConstantsConstants are a special case of the power rule, because $x^0 = 1$. The power rule technically works on contants if you think of them this way, but most students prefer to think of them as their own thing.
Antiderivatives of ConstantsFor any real number constant $a$,$$\int a \, dx = ax + C$$
In words, this tells us that the antiderivative of a lone number is equal to the same number multiplied by $x$ (assuming $x$ is the variable).Linear Operator StuffAs usual, linear operator concepts refer to fairly intuitive things pertaining to coefficients and sums. In short, if you have a sum of power rule type terms, then the integral of the whole expression is equal to the sum of the integral of each term. e.g.$$\int x^2 + \sqrt{x} - \frac{2}{x} \; dx$$$$=\int x^2 \, dx + \int \sqrt{x} \, dx - \int \frac{2}{x} \; dx$$Additionally, constant coefficients attached to variable terms will "stick with" the term through the integration process. e.g.$$\int 5x^3 \, dx = 5 \int x^3 \, dx$$Irrational NumbersJust like with derivatives, teachers will often throw one or two weird questions on a quiz that involve irrational number exponents, like $\sqrt{2}$ or $\pi$.

Example 4Evaluate $\displaystyle \int x^{e} \, dx$.$\blacktriangleright$ The power rule tells us to augment the exponent by $1$, and then divide by the new power. This will not be any different just because the exponent is not an integer.$$\int x^{e} \, dx = \frac{x^{e+1}}{e+1} + C$$

Definite Integration

As we recently learned, integrals with limits must be evaluated using the Fundamental Theorem of Calculus ». Polynomial or power rule terms usually involve a lot of fractions, so it's important to either be careful or to use a calculator.

Example 5$$\int_{0}^{2} 3x^3 -\frac{9x^2}{2} + 7x \, dx$$$\blacktriangleright$ First find the antiderivatives:$$= \Bigg[ \frac{3x^4}{4} - \frac{9x^3}{6} + \frac{7x^2}{2} \; \Bigg\rvert_{0}^{2}$$$$= \Bigg[ \frac{3x^4}{4} - \frac{3x^3}{2} + \frac{7x^2}{2} \; \Bigg\rvert_{0}^{2}$$Finally, evaluate the integral at the integration limits.\begin{align} = & \Bigg[ \frac{3(2)^4}{4} - \frac{3(2)^3}{2} + \frac{7(2)^2}{2} \Bigg] \\ & - \Bigg[ \frac{3(0)^4}{4} - \frac{3(0)^3}{2} + \frac{7(0)^2}{2} \Bigg] \\ \\ = & 14 \end{align}

Put It To The Test

Evaluate each of the following integrals.

Example 6$$\int 3x^8 - 4x^3 + x + 3 \, dx$$
Show solution
$\blacktriangleright$ Make sure you simplify all terms.$$= \frac{3x^9}{9} - \frac{4x^4}{4} + \frac{x^2}{2} + 3x + C$$$$= \frac{x^9}{3} - x^4 + \frac{x^2}{2} + 3x + C$$

Example 7$$\int x^3 + \sqrt[4]{x^2} - \frac{1}{x^3} \, dx$$
Show solution
$\blacktriangleright$ Use fraction and negative exponents to re-write the problem. Simplify along the way where possible.$$\int x^3 + \sqrt[4]{x^2} - \frac{1}{x^3} \, dx$$$$= \int x^3 + x^{1/2} - x^{-3} \, dx$$$$= \frac{x^4}{4} + \frac{x^{3/2}}{3/2} - \frac{x^{-2}}{-2} + C$$$$= \frac{x^4}{4} + \frac{2x^{3/2}}{3} + \frac{1}{2x^2} + C$$

Example 8$$\int x^{\pi - 2} \, dx$$
Show solution
$\blacktriangleright$ While $\pi - 2$ is not an integer, it is a real number, and the power rule applies.$$\int x^{\pi-2} \, dx = \frac{x^{\pi-1}}{\pi-1} + C$$

Example 9$$\int \left(x^2 + 1 \right)^2 \, dx$$
Show solution
$\blacktriangleright$ There isn't really a chain rule for integrals, but this problem isn't too much trouble to multiply out. The challenge is recognizing that you have to distribute in the first place.$$\int \left(x^2 + 1 \right)^2 \, dx = \int \left(x^2 + 1 \right)\left(x^2 + 1 \right) \, dx$$$$= \int x^4 + 2x^2 + 1 \,dx = \frac{x^5}{5} + \frac{2x^3}{3} + x + C$$

Example 10$$\int_{1}^{9} 4x^2 - \frac{1}{\sqrt{x}} \; dx$$
Show solution
$\blacktriangleright$ First, it's helpful to re-write the problem in order to get antiderivatives.$$\int_{1}^{9} 4x^2 - \frac{1}{\sqrt{x}} \; dx$$$$= \int_{1}^{9} 4x^2 - x^{-1/2} \, dx$$$$= \Bigg[ \frac{4x^3}{3} - 2x^{1/2} \Bigg\rvert_{1}^{9}$$Now, evaluate the result with the integration limits.$$= \Bigg[ \frac{4 \cdot 9^3}{3} - 2(9)^{1/2} \Bigg] - \Bigg[ \frac{4 \cdot 1^3}{3} - 2(1)^{1/2} \Bigg]$$$$\frac{2900}{3}$$

Lesson Takeaways
• Understand how the power rule for integration works
• Be familiar with common helpful tips and tricks for roots, radicals, and reciprocals
• Know how to deal with sums and coefficients using linear operator concepts
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