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Integration Techniques 3-By parts

Today, we continue to look at integrals using the different integration techniques shown in Calculus 2. In today’s example we find \(\int_{0}^{1} x^{2}e^{x}dx\).  After determining that the best technique of integration to do this is by using reduction formulas, we then evaluate this integral.

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 \(\int_{0}^{1}x^{2}e^{x}dx\).

Instead of evaluating this definite integral immediately, we will instead find the indefinite integral \(\int x^{2}e^{x}dx\). Then, once we have found an anti-derivative, we will return return to our original problem. In finding \(\int x^{2}e^{x}dx\), we begin by going through the questions in our decision tree to determine the best technique of integration to use.

  • Can you guess the answer?
    I will work under the assumption that we are not able to guess integral.
  • Is there an inside function?
    For this problem, there is no inside function.
  • Is there a product of functions?
    Yes, there is.
  • Is this the product of trig functions?
    No, we do not have trig functions in this problem.

We should therefore try integration by parts to determine what this indefinite integral is. Before we start, we recall that the formula for integration by parts is found using the product rule for derivatives. That is, we have
\begin{align*}
\frac{d}{dx}(FS)&=S \frac{d}{dx}F+F \frac{d}{dx}S \text{ so } \\
\int \frac{d}{dx}(FS) dx&=\int( S \frac{d}{dx}F+F \frac{d}{dx}S)dx \\
(FS)&=\int( S \frac{d}{dx}F)dx +\int(F \frac{d}{dx}S)dx \\
\int F\frac{d}{dx}S dx &=FS-\int S \frac{d}{dx}F dx \text{ or }\\
\int FS’dx&=FS-\int SF’dx.
\end{align*}

If you know the formula for integration by parts, you don’t need to rederive it every time you want to use it. However, I wanted to include this so that you remember where the formula is coming from. Now that we are going to use this formula, we need to assign a \(F\) and \(S’\) from the integral. Note that we want to pick \(S’\) so that it isn’t difficult to integrate. We also want to pick \(F\) so that it isn’t difficult to differentiate. In this case, we can integrate and differentiate both functions, so either choice is viable. We will therefore work through both to see what happens.

Let \(F=e^{x}\) and \(S’=x^{2}\).

In this case we get that \(F’=(e^{x})’=e^{x}\) and \(S=\int x^{2}dx=\frac{x^{3}}{3}\) (note that we will need the constant of integration at the end, but we don’t have to carry it through in each step). We therefore have that
\begin{align*}
\int x^{2}e^{x}dx&=e^{x}\frac{x^{3}}{3}-\int \frac{x^{3}}{3}e^{x}dx.
\end{align*}

As we look at our new integral, we realize that the new integral we have to evaluate is more complicated than our original integral. Therefore, we have actually made the problem more difficult instead of less difficult. If this occurs when using integration by parts, we want to start over using a different choice.

Let \(F=x^{2}\) and \(S’=e^{x}\).

Here we have that \(F’=2x\) and \(S=e^{x}\). We, therefore, get that
\begin{align*}
\int x^{2}e^{x}dx=x^{2}e^{x}-\int e^{x} 2x dx.
\end{align*}

When we look at the new integral we have to calculate, we note that this integral is less complicated than the original because the power of \(x\) is lower than the power of \(x^{2}\). We will now start a new problem finding \(\int 2xe^{x}dx\).

Find \(\int 2xe^{x}dx\).

We again begin by working our way through the decision tree.

  • Can you guess the answer?
    I will work under the assumption that we are not able to guess integral.
  • Is there an inside function?
    For this problem, there is no inside function.
  • Is there a product of functions?
    Yes, there is.
  • Is this the product of trig functions?
    No, we do not have trig functions in this problem.

Our answers therefore lead us to the fact that we will again have to use integration by parts. This time when doing so, we want to make sure that we stay consistent with our choice of \(F\) and \(S’\). That is, if we switch our choices and let \(F=e^{x}\) and \(S’=2x\), we will undo all the work we just did. Hence, we will let \(F=2x\) and \(S’=e^{x}\). We then get \(F’=2\) and \(S=e^{x}\), so
\begin{align*}
\int 2xe^{x}dx=2xe^{x}-\int 2e^{x}dx.
\end{align*}

Again, we are left with a less complicated integral than we started with, but we still have to calculate this. In order to find \(\int 2e^{x}dx\), I would begin by guessing. Since we have an exponential, the integral should look like \(e^{x}\). However, \(\frac{d}{dx}=e^{x} \neq 2e^{x}\). Since we are off by a constant, we can multiply our original guess by this constant and arrive at \(\int 2e^{x}dx=2e^{x}+c\).

Combining this together

Now that we have finished all the needed integrals, we need to combine all of this together. We therefore get that
\begin{align*}
\int x^{2}e^{x}dx&=x^{2}e^{x}-(2xe^{x}-2e^{x})+c \\
&=x^{2}-2xe^{x}+2e^{x}+c.
\end{align*}

Now that we have found the indefinite integral, we can now use one anti-derivative to find the definite integral. That is, we now find that
\begin{align*}
\int_{0}^{1}x^{2}e^{x}dx&=x^{2}e^{x}-2xe^{x}+2e^{x}|_{0}^{1} \\
&=e-2e+2e-(0-0+2) \\
&=e-2.
\end{align*}

We have therefore found the solution to our original problem.

Conclusion

In this problem, we were able to find the correct technique of integration fairly quickly. However, we still had a lot of work to do because the problem required us to use this technique multiple times before we were finally able to arrive at a solution. In this case, it was important to make sure were making progress with each step. If you do make progress, this is an indicator you should keep going. Whereas, if you are seeming to go backwards, it is evidence that you should try something different.

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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