# How to Integrate xe^x: Steps & Tutorial 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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