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# The Method of Substitution - Concept

###### 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.

The substitution method is a very valuable way to evaluate some indefinite integrals. The **substitution method** adds a new function into the one being integrated, and substitutes the new function and its derivative in order to make finding the wanted antiderivative easier.

Let's take a look at a differentiation problem. If I asked you to differentiate one tenth of 1 plus x cubed to the tenth power you'll probably use the chain rule right? One tenth times 10 1 plus x cubed to the ninth times 3x squared. The 3x squared comes from the derivative of the 1 plus x cubed. So this simplifies the one tenth and the tenth cancels so 1 plus x cubed to the ninth times 3x squared. So we use the chain rule to differentiate something like this. But remember every derivative formula can also be written as an integral formula so if I take this as my integrand the integral of this function with respect to x equals this function one tenth, 1 plus x cubed all raised to the tenth power plus c. Now what I want to ask you about this is how would we have done this if we didn't actually start with a derivative problem, if we didn't actually know what the answer was to begin with how would we have integrated this? And the answer is the method of substitution.

We use the chain rule to get this derivative the method of substitution kind of reverses the process of the chain rule it undoes the chain rule. So this is like a reverse chain rule and let me show you how it works. Whenever you use the chain rule or the method of substitution you usually have a composite function of some kind involving your integral. You want to look at the inside part of that composite function and in this case that's the 1 plus x cubed and you're going to substitute for that. This is essentially a change of variables trick. I'm going to let w equal 1 plus x cubed so this is going to become w. And then I need the derivative of that with respect to x.

The derivative of w is going to be 3x squared and very important whenever you have an integral you always have this little dx or d something. This is called the differential you can get a differential from a derivative like this by multiplying both sides by dx, so the differential I'm going to need to change to is dw and so this is going to be my conversion. Now let me show you how this works, so the 1 plus x cubed that's w to the ninth times 3x squared dx that's exactly dw so by a change of variables I've turned this difficult integral into this very easy one. I can integrate this using the power rule for antidifferentiation and remember the way that works is you add 1 to the exponent this becomes w to the tenth and you divide by that same new exponent plus c.

Now because I want my answer to be in terms of x I need to convert back again and remember in the w is 1 plus x cubed I just plug that back in 1 plus x cubed all to the tenth over 10 plus c and that's it. That's how you use the method of substitution to obtain an antiderivative for a complicated function like this but basically undoes the chain rule so whenever you see a composite function or something that you don't think corresponds to any of the integration formulas that you know try to use the method of substitution it works a lot.

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###### 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!”

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