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Section 1-8 : Improper Integrals

2. Determine if the following integral converges or diverges. If the integral converges determine its value.

\[\int_{{ - \infty }}^{0}{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}}\]

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Hint : Don’t forget that we can’t do the integral as long as there is an infinity in one of the limits!
Start Solution

First, we need to recall that we can’t do the integral as long as there is an infinity in one of the limits. Therefore, we’ll need to eliminate the infinity first as follows,

\[\int_{{ - \infty }}^{0}{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}} = \mathop {\lim }\limits_{t \to \, - \infty } \int_{t}^{0}{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}}\]

Note that this step really is needed for these integrals! For some integrals we can use basic logic and “evaluate” at infinity to get the answer. However, many of these kinds of improper integrals can’t be done that way! This is the only way to make sure we can deal with the infinite limit in those cases.

So even if this ends up being one of the integrals in which we can “evaluate” at infinity we need to be in the habit of doing this for those that can’t be done that way.

Show Step 2

Next, let’s do the integral. We’ll not be putting a lot of explanation/detail into the integration process. By this point it is assumed that your integration skills are getting pretty good. If you find your integration skills are a little rusty you should go back and do some practice problems from the appropriate earlier sections.

In this case we need to do integration by parts to evaluate this integral. Here is the integration work.

\[\begin{align*}u & = 1 + 2x & \hspace{0.25in} \to & \hspace{0.25in} & du & = 2dx\\ dv & = {{\bf{e}}^{ - x}}\,dx & \hspace{0.25in} \to & \hspace{0.25in} & v & = - {{\bf{e}}^{ - x}}\end{align*}\] \[\int{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}} = - \left( {1 + 2x} \right){{\bf{e}}^{ - x}} + 2\int{{{{\bf{e}}^{ - x}}\,dx}} = - \left( {1 + 2x} \right){{\bf{e}}^{ - x}} - 2{{\bf{e}}^{ - x}} + c = - \left( {3 + 2x} \right){{\bf{e}}^{ - x}} + c\]

Note that we didn’t do the definite integral here. The limits don’t really affect how we do the integral and so we held off dealing with them until the next step.

Show Step 3

Okay, now let’s take care of the limits on the integral.

\[\int_{{ - \infty }}^{0}{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}} = \mathop {\lim }\limits_{t \to \,\, - \infty } \left. {\left( { - \left( {3 + 2x} \right){{\bf{e}}^{ - x}}} \right)} \right|_t^0 = \mathop {\lim }\limits_{t \to \, - \infty } \left( {\left( {3 + 2t} \right){{\bf{e}}^{ - t}} - 3} \right)\] Show Step 4

We now need to evaluate the limit in our answer from the previous step. In this case we can see that the first term will go to negative infinity since it is just a product of one factor that goes to negative infinity and another factor that goes to infinity. Therefore, the full limit will also be negative infinity since the constant second term won’t affect the final value of the limit.

\[\int_{{ - \infty }}^{0}{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}} = \mathop {\lim }\limits_{t \to \, - \infty } \left( {3 + 2t} \right){{\bf{e}}^{ - t}} - \mathop {\lim }\limits_{t \to \infty } 3 = \left( { - \infty } \right)\left( \infty \right) - 3 = - \infty - 3 = - \infty \] Show Step 5

The final step is to write down the answer!

In this case, the limit we computed in the previous step existed and was negative infinity. Therefore, the integral diverges.