 ###### 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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# Simple Substitutions - 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.

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To integrate a root of a function more complicated than x, one must use the substitution method. The function within the root is the part to substitute 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 take on another problem using integration using integration by substitution. Here I have the integral of the cube root of 6 minus 3x dx. If I were integrating the cube root of x, I wouldn’t need substitution. I could just change that to x to the 1/3 and integrate it using the power rule for antiderivatives. But this is not just the cube root of x. Anything other than x and you really should use the method of substitution.

I’m going to substitute for the inside part. W equals 6 minus 3x. So dw is going to equal -3dx. Now I have a dx but not a -3dx. I’m just going to multiply both sides by -1/3 and just get rid of the -3 here. So -1/3dw equals dx. And now I have something that I can replace the dx with. This becomes -1/3dw, this part becomes the cube root of w and that’s my integral.

I’m going to pull the -1/3 out in front and change this to a power. It’s w to the 1/3. Then I can use the power rule on this, and that’s going to give me w to the 1/3 plus 1. 1/3 plus 1 is 4/3 divided by 4/3, which is the same as multiplying by ¾, but I also have this fraction, the -1/3 in front plus c. The 3s cancel giving me -1/4w to the 4/3 plus c.

Finally I re-substitute for w. 6 minus 3x. So -1/4 (6 minus 3x) to the 4/3 plus c. Now let’s test this by checking using differentiation. I have -1/4 (6 minus 3x) to the 4/3 plus c. When I differentiate this function, the constant pulls out in front, -1/4, the exponent comes down, 4/3, I replace it with 1 less. 1 less than 4/3 is 1/3. And I multiply by derivative of 6 minus 3x and that’s -3. The derivative of plus c is just 0.

So let’s see what I have. -1/4 and 4/3, the 4s cancel and I have -1/3 times -3 is just 1, and I’m left with 6 minus 3x to the 1/3. This is exactly what I started with. Take a look back. The cube root of 6 minus 3x is exactly what I began with. So this answer is correct; -1/4(6 minus 3) to the 4/3 plus c.