# Antiderivatives of Exponentials

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Calculus $\longrightarrow$
Understanding Integration $\longrightarrow$

Objectives
• Using what we know about derivatives, understand how to find the antiderivative of terms of the form $e^{kx}$
• Learn how to integrate exponential terms when the base is not $e$
• Practice taking definite and indefinite integrals of exponential functions
Lesson Description

Leveraging the derivative knowledge we have about exponentials, we will understand how to find antiderivatives of them as well. We'll start with the base case antiderivative of $e^x$, before working on antiderivatives of both $e^{kx}$ and $a^{kx}$, where $a$ and $k$ are real number constants.

Practice Problems

Practice problems and worksheet coming soon!

## More Fun with Exponentials

When we studied exponential functions through the lens of derivatives », we learned that the parent function $f(x) = e^x$ was incredibly unique in that it is the only function whose derivative is itself.As we have begun to see with integration, the key to antiderivatives is looking for the function whose derivative we know - in other words, thinking of derivatives in reverse. Since the derivative of $e^x$ is $e^x$, it should be unsurprising to learn that the antiderivative of $e^x$ is also $e^x$.
Antiderivative of exThe antiderivative of $f(x) = e^x$ is $e^x$.$$\int e^x \, dx = e^x + C$$

The more common task will be to work with functions of the form $f(x) = e^{kx}$$. As with derivatives, the result will involve the k coefficient.Recall that$$\frac{d}{dx} \, e^{kx} = ke^{kx}$$It follows that "going backwards" with an antiderivative will require that we divide by the constant, instead of multiply by it. Antiderivative of ekxFor a function of the form f(x) = e^{kx}:$$\int e^{kx} \, dx = \frac{e^{kx}}{k} + C$$Example 1$$\int e^{7x} \, dx$$\blacktriangleright According to the definition above:$$\int e^{7x} \, dx = \frac{e^{7x}}{7} + C$$## Non-Natural Base Exponentials While we will work with non-natural base exponentials only infrequently, it is important to know how to handle them when they show up. Similarly to derivatives, these integrals require an adjustment factor that involves the natural logarithm of the base.Recall for derivatives that for f(x) = b^x, where b is any real number, that$$\frac{d}{dx} \, b^{x} = b^{x} \cdot \ln(b)$$For antiderivatives, we will need to instead divide by the adjustment factor. Antiderivative of bxFor functions of the form f(x) = b^x, we have$$\int b^x \, dx = \frac{b^x}{\ln(b)} + C$$Example 2Find the antiderivative of 5(3)^x.\blacktriangleright The coefficient 5 can come out of the integration, so, by the above definition, we have$$5 \int 3^x \, dx = \frac{5(3)^x}{\ln(3)} + C$$## Definite Integration - Exponentials Using the Fundamental Theorem of Calculus, we know definite integrals are not very different from indefinite ones - we just have to do some plug-in and arithmetic.You'll find that teachers will often use both integers and logarithm values for integration limits. Example 3$$\int_{-2}^{2} e^{x} \, dx$$\blacktriangleright The antiderivative of e^x is e^x. Then, plug in limits.$$\int_{-2}^{2} e^{x} \, dx = e^x \Bigg\rvert_{-2}^{2}= e^2 - \frac{1}{e^2}$$Example 4$$\int_{\ln(5)}^{5} e^{2x} \, dx\blacktriangleright \; \int_{\ln(5)}^{5} e^{2x} \, dx = \frac{e^{2x}}{2} \; \Bigg\rvert_{\ln(5)}^{5}= \left( e^5 \right) - \left( e^{2\ln(5)} \right)$$In order to fully simplify for full credit, we must realize that 2\ln(5) is equal to \ln(25), using the power rule of logarithms.$$\longrightarrow e^5 - e^{\ln(25)}=e^5 - 25$$## Mr. Math Makes It Mean The particularly annoying questions are the ones where we have to integrate non-natural number exponentials that also have constants in the exponent. Just put both ideas together by keeping track of each rule separately if need be. Example 5$$\int 6^{4x} \, dx\blacktriangleright = \frac{6^{4x}}{4\ln(6)} + C$$The 4 in the denominator comes from the rule that governs integrating an exponential to the power kx. The \ln(6) term comes from the rule that governs integrating an exponential other than base e. Pro Tip Be aware that on multiple choice questions on the AP exam, they like to do this because the denominator of the correct answer choice could also be \ln(1296), since 4\ln(6) is equivalent to \ln(6^4). Also, when non-natural bases are in play, teachers like to include log based integration limits. Example 6$$\int_{0}^{\log_5 (10)} 5^{x} \, dx$$\blacktriangleright When we get to the step where we plug in the limits, we need to simplify completely.$$\int_{0}^{\log_5 (10)} 5^{x} \, dx = \frac{5^x}{\ln(5)} \; \Bigg \rvert_{0}^{\log_5 (10)}= \frac{1}{\ln(5)} \; \left[ 5^{log_5 (10)} - 5^0 \right]= \frac{1}{\ln(5)} (10 - 1)= \frac{9}{\ln(5)}$$## Put It To The Test Example 7$$\int 2e^{2x} \, dx$$Show solution$$\blacktriangleright \; e^{2x} + C$$Example 8Prove that$$\int e^{10x} \, dx = \frac{e^{10x}}{10} + C$$by taking the derivative of the result. Show solution \blacktriangleright Use the derivative rule for exponentials.$$\frac{d}{dx} \; e^{kx} = ke^{kx}$$Therefore$$\frac{d}{dx} \; \frac{1}{10} \cdot e^{10x}+C  = \frac{1}{10} \cdot 10e^{10x} = e^{10x}$$The +C "disappeared" because C is a constant, so its derivative is zero. Example 9$$\int -2 \cdot 3^{-x} \, dx $$Show solution \blacktriangleright Think of the negative exponent as a coefficient of -1.$$\int -2 \cdot 3^{-x} \, dx = -2 \cdot \frac{3^{-x}}{(-1)\ln(3)} +C= \frac{2(3^{-x})}{\ln(3)} + C$$Example 10$$\int 6^{2x/3} \, dx$$Show solution \blacktriangleright Two rules need to be followed. First, the coefficient in the exponent is 2/3, which we're told to divide by when we integrate. Second, the exponential base is not e, so we'll need to include the natural log of 6 in the denominator of our answer as well.Recall that dividing by 2/3 is the same as multiplying by 3/2, and we'll have our answer:$$\int 6^{2x/3} \, dx = \frac{3 \cdot 6^{2x/3}}{2\ln(6)} + C$$Note that in a multiple choice setting, the correct answer could have \ln(36) in the denominator instead, since 2\ln(6)=\ln(36). Example 11$$\int 4\cdot 4^{x-1} \, dx$$Show solution \blacktriangleright We did not discuss how to work with linear transformations (ax+b instead of x) in the exponent, but we can use the algebra rules of exponents to re-write this as something we know better how to work with:$$4^{x-1} = 4^x \cdot 4^{-1} = \frac{4^x}{4}$$And so,$$\int 4\cdot 4^{x-1} \, dx = \int \frac{4}{4} \cdot 4^x \, dx= \int 4^x \, dx$$We now know how to deal with it.$$\int 4^x \, dx = \frac{4^x}{4} + C$$Which, possibly unsurprisingly, could actually be written as$$\int 4^x \, dx = 4^{x-1} + C$$I Used To Know That! Core algebra exponent rules will never stop being useful:$$x^{a} \cdot x^{b} = x^{a+b}\frac{x^{a}}{x^{b}} = x^{a-b}\left(x^a\right)^b = x^{ab}$$Example 12$$\int_{\ln(3)}^{\ln(10)} e^{2x} \, dx$$Show solution \blacktriangleright$$$\int_{\ln(3)}^{\ln(10)} e^{2x} \, dx = \frac{e^{2x}}{2} \; \Bigg \rvert_{\ln(3)}^{\ln(10)}= \frac{1}{2} \left[ \left( e^{2\ln(10)} \right) - \left( e^{2\ln(3)} \right) \right]= \frac{1}{2} \left[ \left( e^{\ln(100)} \right) - \left( e^{\ln(9)} \right) \right]= \frac{1}{2} \; \left[ \left( 100 \right) - \left( 9 \right) \right] = \frac{91}{2}$$Example 13$$\int_{\log_5 (2)}^{1} 2 \cdot 5^{x} \, dx$$Show solution \blacktriangleright$$$\int_{\log_5 (2)}^{1} 2 \cdot 5^{x} \, dx = \frac{2 \cdot 5^{x}}{\ln(5)} \; \Bigg \rvert_{\log_5 (2)}^{1} = \frac{2}{\ln(5)} \left[ \left( 5^1 \right) - \left( 5^{\log_5 (2)} \right) \right] = \frac{2}{\ln(5)} \; \left( 5 - 2 \right)= \frac{6}{\ln(5)}$$Example 14$$\int_{\log_4 (9)}^{3} 6 \cdot 2^{-x} \, dx$$Show solution \blacktriangleright$$$ \longrightarrow \frac{-6 \cdot 2^{-x}}{\ln(2)} \; \Bigg \rvert_{\log_4 (9)}^{3}= \frac{-6}{\ln(2)} \; \left[ \left( 2^{-3}\right) - \left( 2^{-\log_4 (9)}\right) \right]$$Depending on your situation, you may be able to leave your answer in this form. However, many teachers want you to fuss with the cleanup on stuff like this, because that right-most term simplifies greatly:$$2^{-\log_4 (9)} = \left( 4^{1/2} \right)^{-\log_4 (9)}= 4^{-\frac{1}{2}} \log_4 (9) = 4^{-\log_4 (3)}= -\frac{1}{3}$$So overall,$$\int_{\log_4 (9)}^{3} 6 \cdot 2^{-x} \, dx= \frac{-6}{\ln(2)} \left[ \frac{1}{8} - \frac{1}{3} \right]= \frac{-6}{\ln(2)} \; \left( -\frac{5}{24} \right)= \frac{5}{4\ln(2)}$$Or,$$\frac{5}{\ln(16)}

Lesson Takeaways
• Know the basic principles of integrating exponential functions
• Learn how to adjust if asked about exponential antiderivatives of functions other than base $e$
• Practice definite and indefinite integration with exponentials
• Understand when, why, and how to simplify answers to both definite and indefinite exponential integrals
• Learn common complications that teachers use on tests for this topic
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