Integrals of Exponential Functions

Find the derivative of $f\left(x\right)=e^{2x^2}$.

Let $u=2x^2$and differentiate using The Chain Rule.

$\frac{df}{dx}=\frac{d}{du}{e}^u\frac{du}{dx}$

Differentiate the exponential function.

$\frac{df}{dx}=e^u\frac{du}{dx}$

Use The Power Rule to differentiate $u=2x^2$.

$\frac{du}{dx}=4x$

Substitute back $u=2x^2$and$\frac{du}{dx}=4x$.

$\frac{df}{dx}=e^{2x^2}4x$

Rearrange the expression.

$\frac{df}{dx}=4x{e}^{2x^2}$

Evaluate the integral $\int e^{3x}dx$.

Since the argument of the exponential function is 3x, we need to do Integration by Substitution.

Let $u=3x$. Find du using The Power Rule.

$u=3x\to \frac{du}{dx}=3$

$\frac{du}{dx}=3\to du=3dx$

Isolate dx.

$dx=\frac{1}{3}du$

Substitute $u=3x$ and $dx=\frac{1}{3}du$ in the integral.

$\int e^{3x}dx=\int e^u\frac{1}{3}du$

Rearrange the integral.

$\int e^{3x}=\frac{1}{3}\int e^{u}du$

Integrate the exponential function.

$\int e^{3x}dx=\frac{1}{3}{e}^u+C$

Substitute back $u=3x$ in the integral.

$\int e^{3x}dx=\frac{1}{3}{e}^{3x}+C$

Evaluate the definite integral ${\int }_0^1{e}^{x}dx$.

Find the antiderivative of $e^x$.

$\int e^x=e^x+C$

Use The Fundamental Theorem of Calculus to evaluate the definite integral.

${\int }_0^1{e}^{x}dx={\overline{)\left(e^x+C\right)}}_0^1$

${\int }_0^1{e}^{x}dx=\left(e^1+C\right)-\left(e^0+C\right)$

Use the properties of exponents and simplify.

${\int }_0^1{e}^{x}dx=e-1$

Up to this point, we have an exact result. You can always use a calculator if you need to know the integral's numerical value.

Use a calculator to find the numerical value of the definite integral.

${\int }_0^1{e}^{x}dx=1.718281828...$

Evaluate the definite integral ${\int }_0^{\infty }{e}^{-2x}dx$.

Begin by finding the antiderivative of the given function.

Let $u=-2x$. Find du using The Power Rule.

$u=-2x\to \frac{du}{dx}=-2$

$\frac{du}{dx}=-2\to du=-2dx$

Isolate $dx$.

$dx=-\frac{1}{2}du$

Substitute $u=-2x$ and$dx=-\frac{1}{2}du$in the integral.

$\int e^{-2x}dx=\int e^u\left(-\frac{1}{2}du\right)$

Rearrange the integral.

$\int e^{-2x}dx=-\frac{1}{2}\int e^{u}du$

Integrate the exponential function.

$\int e^{-2x}dx=-\frac{1}{2}{e}^u+C$

Substitute back $u=-2x$.

$\int e^{-2x}dx=-\frac{1}{2}{e}^{-2x}+C$

In order to evaluate the improper integral, we use The Fundamental Theorem of Calculus, but we evaluate the upper limit as it goes to infinity. That is, we let \(b\rightarrow\infty\) in the upper integration limit.

${\int }_0^{\infty }{e}^{-2x}dx=\underset{b\to \infty }{\lim }\left(-\frac{1}{2}{e}^{-2b}+C\right)-\left(-\frac{1}{2}{e}^{-2\left(0\right)}+C\right)$

Simplify using the Properties of Limits.

${\int }_0^{\infty }{e}^{-2x}dx=-\frac{1}{2}\left(\underset{b\to \infty }{\lim }{e}^{-2b}-e^0\right)$

As \(b\) goes to infinity, the argument of the exponential function goes to negative infinity, so we can use the following limit:

$\underset{x\to \infty }{\lim }{e}^{-x}=0$

We also note that $e^0=1$. Knowing this, we can find the value of our integral.

Evaluate the limit as $b\to \infty$and substitute $e^0=1$.

${\int }_0^{\infty }{e}^{-2x}dx=-\frac{1}{2}\left(0-1\right)$

Simplify.

${\int }_0^{\infty }{e}^{-2x}dx=\frac{1}{2}$

Evaluate the integral $\int 2x{e}^{x^2}dx$.

Note that this integral involves $x^2$ and $2x$in the integrand. Since these two expressions are related by a derivative, we will do Integration by Substitution.

Let $u=x^2$. Find $du$using The Power Rule.

$u=x^2\to \frac{du}{dx}=2x$

$\frac{du}{dx}=2x\to du=2xdx$

Rearrange the integral.

$\int 2x{e}^{x^2}dx=\int e^{x^2}(2xdx)$

Substitute $u=x^2$and $du=2xdx$in the integral.

$\int 2x{e}^{x^2}dx=\int e^{u}du$

Integrate the exponential function.

$\int 2x{e}^{x^2}dx=e^u+C$

Substitute back $u=x^2$.

$\int 2x{e}^{x^2}dx=e^{x^2}+C$

Evaluate the integral $\int (x^2+3x)e^{x}dx$

Use LIATE to make an appropriate choice of u and dv.

$u=x^2+3x$

$dv=e^{x}dx$

Use The Power Rule to find du.

$du=\left(2x+3\right)dx$

Integrate the exponential function to find v.

$v=\int e^{x}dx=e^x$

Use the Integration by Parts formula $\int udv=uv-\int vdu$

$\int (x^2+3x)e^{x}dx=(x^2+3x)e^x-\int e^x(2x+3)dx$

The resulting integral on the right-hand side of the equation can also be done by Integration by Parts. We shall focus on evaluating $\int e^x(2x+3)dx$to avoid any confusion.

Use LIATE to make an appropriate choice of u and dv.

$u=2x+3$

$dv=e^{x}dx$

Use The Power Rule to find du.

$du=2dx$

Integrate the exponential function to find v.

$v=\int e^{x}dx=e^x$

Use the Integration by Parts formula.

$\int e^x(2x+3)dx=(2x+3)e^x-\int e^x(2dx)$

Integrate the exponential function.

$\int e^x(2x+3)dx=(2x+3)e^x-2e^x$

Substitute back the above integral into the original integral and add the integration constant C.

$\int (x^2+3x)e^{x}dx=(x^2+3x)e^x-\left[(2x+3)e^x-2e^x\right]+C$

Simplify by factoring out $e^x$.

$\int (x^2+3x)e^{3x}dx=e^x(x^2+x-1)+C$

Evaluate the integral ${\int }_1^2{e}^{-4x}dx$.

Begin by finding the antiderivative of the function. Then we can evaluate the definite integral using The Fundamental Theorem of Calculus.

Integrate the exponential function.

$\int e^{-4x}dx=-\frac{1}{4}{e}^{-4x}+C$

Use The Fundamental Theorem of Calculus to evaluate the definite integral.

${\int }_1^2{e}^{-4x}dx={\overline{)\left(-\frac{1}{4}{e}^{-4x}+C\right)}}_1^2$

${\int }_1^2{e}^{-4x}dx=\left(-\frac{1}{4}{e}^{-4\left(2\right)}+C\right)-\left(-\frac{1}{4}{e}^{-4\left(1\right)}+C\right)$

Simplify.

${\int }_1^2{e}^{-4x}dx=-\frac{1}{4}\left(e^{-8}-e^{-4}\right)$

Use the properties of exponents to further simplify the expression.

${\int }_1^2{e}^{-4x}dx=\frac{e^{-4}-e^{-8}}{4}$

${\int }_1^2{e}^{-4x}dx=\frac{e^{-8}(e^4-1)}{4}$

${\int }_1^2{e}^{-4x}dx=\frac{e^4-1}{e^8}$

We have seen a shortcut for integrating exponential functions when their argument is a multiple of x.

$\int e^{ax}dx=\frac{1}{a}{e}^{ax}+C$

This saves us plenty of time for sure! However, one common mistake is multiplying by the constant rather than dividing.

$\int e^{ax}dx\ne a{e}^{ax}+C$

This might happen to you if you just differentiated an exponential function, maybe you were doing Integration by Parts.

Another common mistake when integrating (not only exponential functions!) is forgetting to add the integration constant. That is, forgetting to add +C at the end of the antiderivative.

Always make sure to add +C at the end of an antiderivative!

$\int e^{x}dx=e^x+C$