How to Integrate xe^x: Steps & Tutorial

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  • 0:00 Solve Using the Product Rule
  • 1:40 Integration by Parts
  • 2:38 The Other Choice for…
  • 3:38 A Reduction Formula
  • 5:49 Lesson Summary
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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we use the product rule and integration by parts to find the integral of xe^x. The natural extension of integration by parts leads to a reduction formula which elegantly extends the integration results.

Solving Using the Product Rule

Let's start by differentiating xex using the product rule. According to this rule, we take the derivative of the first function multiplied with the second function and add the first function multiplied with the derivative of the second. This sounds more complicated than it is, so let's look at an example:


differential


On the right-hand side, the derivative of x is 1 and the derivative of ex is ex.

To clean up the right-hand side:


differential_expanded


Now, we integrate both sides:


integrating


On the left-hand side, the integral 'undoes' the derivative, so the integral of d(xex) is xex.

And, on the right-hand side, the integral of the sum is the sum of the integrals:


integration_expanded


This next part is a re-ordering of the three terms. We are looking for the integral of xex so we place this term on the left-hand side by itself.


re-ordering


(In the next section, we will refer to this equation as the integration by parts formula.)

The integral of ex is just ex.


solving


The constant 'C' is appended to the answer because this is an indefinite integral with unspecified limits of integration. And we are done!

Of course, we could factor out the ex:


simplifying


Now to check this answer, we differentiate and the result should be just xex.

Once again, we use the product rule:


checking


and then expand and simplify:


checking_expanded_done


The answer checks out!

Integration by Parts

So we now know what the answer is. What we've actually done by integrating the expansion of the product rule is to derive the integration by parts formula:


integration_by_parts


The key is selecting a u and dv which will reduce the complexity of the problem by giving us a simpler integral to solve. A good choice is:


a_good_choice_for_u_and_v


For u = x, du = dx. For dv = ex dx, v = ex.

Substituting u, dv, v and du into the integration by parts formula:


substituting_into_the_IBP


Now simplifying, we have the following:


simplifying_IBP


This gives us the same answer that we found before.

The Other Choice for u and dv

Let's explore what would have happened had we made the other choice for u and dv. This will actually lead to an interesting result.

Instead of u = x, we choose u = ex. Also, instead of dv = ex dx, we choose dv = x dx.

Thus, we now have du = ex dx and v = x2/2. Summarizing this other choice:


simplifying_IBP


Note the left-hand side of the integration by parts formula is the same, but the right-hand side has an integral, which has increased in complexity.


IBP_with_other_choice


But the integral on the left-hand side is our friend ex (x - 1):


sustituting_LHS


Thus, we can multiply by 2 and isolate the more complicated integral on the left-hand side:


simplifying_IBP


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