# Integration Techniques 6-Improper Integrals

Today, we continue to look at integrals using the different integration techniques shown in Calculus 2. In this problem, we will find the $$\int_{0}^{\infty} e^{-x}dx$$. In this case we have an improper integral because $$\infty$$ is not a real number, and is, therefore, not in the domain of the function.

Hopefully you will join me each day as I go through these integrals in both a post and in a video. Along the way, I hope you learn something, and, if you are taking Calculus 2, you are in a great position to do very well on any upcoming exams you may have. You can find more Calculus 2 study help posts here. I also have a decision tree available so that you can follow along as we determine what technique of integration to use. I also have a useful list of formula sheet available.

## Find $$\int_{0}^{\infty}e^{-x}dx$$.

As we start this problem, we note that we cannot use the fundamental theorem. In order to use the fundamental theorem, the integrand must be defined over the interval of integration. While the domain of $$e^{-x}$$ is all real numbers, we wouldn’t consider it to be defined at $$\infty$$ since $$\infty$$ is not a real number.

We should note here, that without context, the question itself does not make sense, because we need a number as a limit of integration. We, therefore, have to look more closely at the details to realize that what the problem is really trying to get us to find is the area under $$e^{-x}$$ on an interval starting at $$0$$ and ending at an arbitrarily large value. That is, we define
\begin{align*}
\int_{0}^{\infty}e^{-x}dx=\lim_{b \to \infty}\int_{0}^{b}e^{-x}dx.
\end{align*}

In this case, since $$b$$ is a real number, we can use the fundamental theorem. In order to do so, we will have to use the find the anti-derivative of $$e*{-x}$$. As we do this, we again go through our decision tree.

• Can you guess the answer?
The integrand is an exponential function, so I will guess that the anti-derivative is the original function.
• Take the derivative, is this what we started with?
Note that the derivative of $$e^{-x}$$ is $$-e^{-x}$$, so they are not the same.
• Are you off by a constant?
Yes, so I will multiply the original by the constant we are off by, that is, we let our new guess be $$-e^{-x}$$.
• Take the derivative, is the answer the same as the original?
Yes, the derivative of $$-e^{-x}$$ is $$e^{-x}$$.<\li>We have now found the anti-derivative by guessing and checking, so we can now calculate our improper integral.
\begin{align*}
\int_{0}^{\infty}e^{-x}dx&=\lim_{b \to \infty} \int_{0}^{b}e^{-x}dx \\
&=\lim_{b \to \infty}-e^{-x}|_{0}^{b} \\
&=\lim_{b \to \infty}-e^{-b}-(-e^{-0}) \\
&=\lim_{b \to \infty}\frac{-1}{e^{b}}+1 \\
&=0+1=1.
\end{align*}Hence, we have that the total area under the curve $$e^{-x}$$ on the interval from $$0$$ to $$\infty$$ will be $$1$$. That is, as the right hand limit gets arbitrarily large, the total area approaches 1.

## Conclusion

In this problem, we had some initial difficulty because the integral had $$\infty$$ as a limit. Another way we can have these improper integrals is if the function is undefined at some point in the interval. In these cases, we have to recall that the way these were defined was by taking an integral to an arbitrary point, then finding the limit as we approached the given value. Once we saw this, we could find the integral and limit as we have done in previous examples.

I hope you learned something from this post and found it helpful.  If you did, make sure to share it and the video with anyone else you know that may need some extra help with integrals.

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