Evaluate each definite integral.

$$\int_{1}^{8} \frac{3-y^{1 / 3}}{y^{2 / 3}} d y$$

$\frac{9}{2}$

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Numerade Educator

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

All right. Here we have the definite integral one to eight of the function. Three minus. Why? To the one third all over? Why? To two thirds do I. And so let's look at this numerator here. Three minus y to the one third. If I differentiate this up to a constant, I'm going to get exactly why do the minus two thirds, which is this do you want? So that's telling me I'm gonna let you be three minus y to the one third. Do you hear is going to be okay. Ah, minus one third. Why? To the negative two thirds t. Why? And this is exactly what I have left. Well, I just need to figure out the price. So I have negative three years. I communicate a pickup. A factor of negative three. Why? To the negative two thirds d y. And so this integral to be equal, Tio Well, when y is equal to one, you is equal to two. Looks like three minus one to the one third into the wonders to sew three. One is two is one. When more so use one When? Why is eight and then we have our factor of minus three. From the substitution we have you and then do you? So this should be straightforward. Negative three. Then we just have an anti derivative squared over two evaluated from two to one. So this is negative. Three over, too. One squared, minus two squared. So what did we get? This is gonna be negative three nine over.

Georgia Southern University