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.

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Computing Definite Integrals using Substitution - Problem 3

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.

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When finding the definite integral of a function, you may need to use substitution in order to first find the indefinite integral. Recall that for the method of substitution, replace an expression of x with w. For expressions involving a cube root, typically replace Find its derivative to find dw. Then, integrate as usual with respect to w.

After integrating, replace the w's with the appropriate expression of x. Then, plug in the limits of integration and subtract (if taking the integral from a to b, and the result of the integral is F(x), the result will be F(b) - F(a)).

Let’s solve one more definite integral using the method of substitution. I’m asked to solve the integral from 7 to 35 over 1/3(2x minus 6) to the 1/3. First of all this is one of those substitutions where I have a radical inside the integral. Raising to the 1/3 is the same as taking the cube root of something. So this is 2x minus 6 and the cube root of that.

So I want to substitute for this. W equals 2x minus 6. Dw is going to be just 2dx. I don’t have a 2dx, so I’m going to need to divide both sides by 2. I’ll have ½dw equals dx. I also want to substitute for the limits. So let's start with x equals 7.

X equals 7 becomes what? 2 times 7, 14 minus 6, 8, w equals 8. What about x equals 35? First, I double it, 70 minus 6, 64. So these are my new limits 8 and 64. So I’ll be integrating from 8 to 64. I still have the 1/3. This becomes w to the 1/3 and the dx is ½dw.

I can combine these two constants to 1/6, and integrate from 8 to 64 w to the 1/3 dw. So what’s an antiderivative for w to the 1/3? Remember you just add 1 to the exponent, so 1/3 plus 1 is 4/3 and then I’m going to divide by 4/3 which is the same as multiplying by ¾.

So I’m going to have 1/6 times ¾w to the 4/3. I’m going to evaluate that from 8 to 64. This fraction simplifies a little bit. The 3 and the 6 cancel leaving a 2 in the denominator, so this will become 1/8. 1/8w to the 4/3, from 8 to 64.

What happens when I plug 64 in here? Let me just write it out. 1/8(64) to the 4/3 minus 1/8(8) to the 4/3. So 64 if I look at the exponent here, I have a 3 in the denominator that’s like taking the cube root. This is a perfect cube 64 is 4 cubed. So the cube root of 64 is 4, and then I raise the result to the fourth power. That’s what it means to take the 4/3 power something cube root and then raise to the 4th power.

4 to the 4th is 256. So I have 256 times 1/8 and now what about the other one? 8 to the 4/3, 8 is also a perfect cube so the cube root of that Is 2. Raise that to the fourth power you get 16.1/8 times 16. I know it’s tempting to multiply these out first but notice you have a 1/8 here and here. It’s really 1/8 times the difference 240.

And what is 1/8 times 240? 30, that’s my final answer 30. 30 is the exact value of the integral from 7 to 35 of 1/3(2x minus 6) to the 1/3 dx. Now remember very easy to use method of substitution on a definite integral. Just make sure you substitute for all the x’s inside and on the limits of integration too.

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