##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Substitutions Involving e^x or ln(x) - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

To take the integral of functions involving e raised to some power, substitute the power that e is raised to with w. Then, use the substitution method: find dw by taking the derivative of w with respect to x, substitute the x's in the integrand with the expression for w, and integrate as usual. Replace the w's with the expression for x.

Let's do a problem that involves a method of substitution on an exponential function. Here we're integrating e to the 3 plus 0.4x. We saw before that we can't just immediately apply the integration formula. If you will follow me over here, the integral of e to x dx equals e to the x plus c directly. We actually have to get our integral into this simple form first. The way we do that is by the method of substitution.

So I'm going to substitute for this 3 plus 0.4x. I'm going to turn into the single variable w, so I can use that formula. W equals 3 plus 0.4x. Remember that I have to differentiate this. I need to get a dw for my integral, so dw equals 0.4dx.

Now I don't actually have a 0.4dx. I just have a dx, so I have to divide both sides by 0.4. So 1 over 0.4dw is going to equal dx. What's 1 over 0.4. 1 over 0.4 is the same as 10 over 4. Multiply the top, and bottom by 10. That's the same as 5/2, or 2.5. I'll write 5/2. So that means I can replace dx with 5/2dw. So 5/2dw goes here. This becomes e to the w. I can pull the 5/2 out of the integral. So the 5/2 comes out, and I have the integral of e to the w, dw.

Of course, now I can use the formula. Integral of e to the w is e to the w plus c. So this becomes 5/2e to the w plus c. The last thing I need to do is re-substitute for w. I know w equals 3 plus 0.4x. So I plug that back in. E to the 3 plus 0.4x plus c. That's my final answer for the antiderivative of e to the plus 0.4x. This 5/2 here is necessary. It's like reversing the chain rule, what I've just done.

Now let me check this answer by differentiating it. If I differentiate it, I should get this back again. So let me try that. The derivative with respect to x of 5/2e to 3 plus 0.4x plus c is 5/2. The 5/2, constant can be pulled out, times e to the 3 plus 0.4x, times the derivative of 3 plus 0.4x. The derivative of the inside function, this is the chain rule, and that's 0.4.

Now just referring back to the calculation I did before, 5/2 is the same as 1 over 0.4. This is the same as 1 over 0.4. This will actually then cancel. Indeed I do get e to the 3 plus 0.4x back again, the original function. So that means that this answer is correct. This is the correct antiderivative.

Let's do another example that also involves this same integration formula; e to the x. I have something that doesn't look exactly the same as what I just integrated, but it still has an e to the something. So remembering that I use the method of substitution, whenever I see a composite function, I look to the inside function. 2x³, I'm going to substitute for that. The hope is that I'll get an integral that involves just e to the w.

Well what's dw in this case? Dw is going to be the 3 comes down, and I get 6x²dx, don't forget this dx, because you want to eventually substitute for this dx. So it has to appear in one of your substitution equations.

Now I have an x², and a dx. But I don't have the 6x²dx, so I can just divide both sides by 6, and get 1/6dw equals x²dx. Now I can replace these two things with the 1/6dw. So let me do that.

1/6dw, and the e to the 2x³, that's going to be e to the w. So this worked really nicely, e to the w times 1/6dw. I can pull the 1/6 out and now I can just use this formula. I have it exactly in the form I need. E to a variable. E to the w dw. So the integral is going to be e to the w. 1/6e to the w plus c. All I have to do now is replace the 2x³ in for w. So I get 1/6e to the 2x³, don't forget the plus c is down here. So this is my antiderivative for x²e to the 2x³.

Now I should check this using antidifferentiation. If I antidifferentiate my answer, this guy, I should get the original function back again. So let's see if that happens. 1/6 e to the 2x³ plus e. So the 1/6 pulls out the derivative, and then I differentiate e to the 2x³. That's going to be e to the 2x³, times the derivative of 2x³, which is 6x². Now 6x² times 1/6 is just x². These guys cancel, and that's exactly what I hoped for; x²e to the 2x³ is exactly what I started with. That means that this answer is indeed correct.

So method of substitution; turn this complicated integral into an integral that was exactly like one of my formulas. That's basically what the method of substitution is for. You want to convert an unpleasant looking integral into an integral that looks exactly like one of your formulas.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete