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Numerade Educator



Problem 31 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int^{1}_{0} x \bigl( \sqrt[3]{x} + \sqrt[4]{x} \bigr) \,dx $




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Video Transcript

So we're doing the integral from 0 to 1 of this function. And pointing out the product here um because we don't have a product rule for anti derivatives and instead of writing the cuBA and the 1/4 root, I'm writing what's equivalent to that, which is to the one third powers of cuba at the 1/4 powers is four through. Um And what I would have my sense to is distribute this in. So now it is uh polynomial. And just remember that when you're multiplying with the same base, you add the exponents before thirds And then this would be to the five force power. So now we're ready to do the anti directive which is adding one to the exponent. It might make sense to write it as three thirds to help you add one, I don't know, I just assumed you knew how to do that before and then multiply by the reciprocal. So this one you're adding four force, which is exactly what it did over here. So it's really the same math. But then multiply by the reciprocal. I should be putting equals in here by the way. Um and we're going from 0-1. So it's really nice about this. Problem is when you plug in the upper bound and for your exes well one to any power is just one. Think of the cube root of one is one to the seventh hour still one. So we're just looking at 3/7 plus 4/9. And then you'll be subtracting off. And I still encourage all my students to put in zeros in there. Just because when you do like co sign of zero or either the zero power, you actually get numerical value. So this case you do not get anything subtracting zero is not going to change. So I would get the same denominator in these two. And what I would have to do is multiply this piece by 9/9 to get that denominator of 63. So we're looking at 27. And over here I'd have to multiply by 7/7s To get 28. Um as that X numerator, Uh when you add those, you get the correct answer of 55/63. There you go.