Integrals of exponential functions

$e^x$ stays $e^x$ whether you differentiate or integrate. For base $a$, differentiation multiplies by $\ln a$, so integration divides by it.

When the exponent is linear, as in $e^{2x}$, remember to divide by the coefficient of $x$.

Examples

1. 01What is the value?$\displaystyle \int_{0}^{1} e^{x}\,dx$AnswerHide

   Answer$\displaystyle e-1$

   Steps

   1. $e^{x}$ integrates to $e^{x}$; subtract the lower end from the upper
   2. $\displaystyle e^{1}-e^{0}$$\displaystyle = e-1$
   3. We subtract $e^{0}=1$, so the answer is not $e$ itself
2. 02What goes in □?$\displaystyle \int a^{x}\,dx = \frac{a^{x}}{\square}$AnswerHide

   Answer$\displaystyle \log_e a$

   Steps

   1. Differentiating $a^{x}$ multiplies by $\log_e a$; integrating divides by it instead
   2. $\displaystyle \frac{a^{x}\log_e a}{\log_e a}$$\displaystyle = a^{x}$
   3. That cancels back, so $\square$ is $\log_e a$
3. 03What is this?$\displaystyle \int e^{x}\,dx$AnswerHide

   Answer$\displaystyle e^{x}+C$

   Steps

   1. $e^{x}$ is the function that differentiating leaves alone, so integrating leaves it alone too
   2. $\displaystyle f(x)$$\displaystyle = e^{x}$$\displaystyle f'(x)$$\displaystyle = e^{x}$
   3. The power rule does not apply: in $e^{x}$ the $x$ is the exponent, not the base
4. 04Integrate$\displaystyle \int e^x \, dx$AnswerHide

   Answer$\displaystyle e^x + C$
5. 05What is this?$\displaystyle \int 3^{x}\,dx$AnswerHide

   Answer$\displaystyle \frac{3^{x}}{\log_e 3}+C$

   Steps

   1. Differentiating $3^{x}$ brings out a factor of $\log_e 3$
   2. $\displaystyle f(x)$$\displaystyle = 3^{x}$$\displaystyle f'(x)$$\displaystyle = 3^{x}\log_e 3$
   3. Integration runs the other way, so divide by that factor in advance
   4. $\displaystyle \frac{3^{x}\log_e 3}{\log_e 3}$$\displaystyle = 3^{x}$
   5. It cancels and we are back. Differentiating multiplies; integrating divides
6. 06What is this?$\displaystyle \int 5^{x}\,dx$AnswerHide

   Answer$\displaystyle \frac{5^{x}}{\log_e 5}+C$

   Steps

   1. Differentiating $5^{x}$ brings out a factor of $\log_e 5$
   2. $\displaystyle f(x)$$\displaystyle = 5^{x}$$\displaystyle f'(x)$$\displaystyle = 5^{x}\log_e 5$
   3. Integration runs the other way, so divide by that factor in advance
   4. $\displaystyle \frac{5^{x}\log_e 5}{\log_e 5}$$\displaystyle = 5^{x}$
   5. It cancels and we are back. Differentiating multiplies; integrating divides

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