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Solve the given problems by integration.Find the volume generated when the first-quadrant area bounded by $y=e^{x}$ and $x=2$ is rotated about the $x$ -axis. [Hint: After setting up the integral, rewrite $\left.\left(e^{x}\right)^{2} \text { as }\left(e^{x}\right)\left(e^{x}\right) .\right]$

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 1

The General Power Formula

Integrals

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in this question, we want to find the volume generated by this curve And bounded by as equals to two and rotated about the excesses. In general when we have a cough White was two epics. And we want to find the volume from Mexico to A to be generated about the excesses over here. We'll just get a small strip over here. The week is so small. We'll just call it D. X. Okay. And of course the heart here is why now as he generates over the excesses is creating a circular disk. The circular, jeez the wife is the X. Or the height of the disease dX. And the radius is why. So the volume of this. This is pie our square heat, right pi y square the X. No. To get the entire volume from A to B. We would just sum up now since this is a continuous submission. So it will be an integration sign from A to B. Hi y square the X. You can leave the pie outside. And so that would be your volume. So to find a volume for this so volume equals two pi summing from 0 to 2. Why square the X. Yeah. So that will be E. To the X times X. Because that's why square E. X. There are a couple of ways to integrate this. But for this topic we're using substitution method and integration power rule. So for substitution method licks you'll be eating power X. The U. Over the X. Is it a power X. So the you is it apart X. Dx. So this part over here is the you now we need to substitute these X values here as well. We used values. So when X equals to the lower limit zero You is either power zero and there is one When X is the upper limit of two. U.S.E. No, we also need the power rule. The power rule says that when we integrate you to power and we respect to you, we will just get you to the power and plus one. So we add one to the power and divide by the new power. Now I put classy here because this is an indefinite integral but for definite integral we do not need to put A plus E. So now we are ready to substitute soap. I now instead of X equals zero, it will be you close to one and for the upper limit of X equals two, you'll be U equals to the square. Now it'll power X over here is you and it'll power X dx over here is the you. So now we're ready to integrate, this is you to power one. Using Power rule, it will become you to power one plus one over the new power one plus one and is within the epitome of e square and lower the MIr one mm. Now I can bring this two Outside so it will be pi over two. Yeah, you square. So you square. I'll be subbing the upper limit in first, so will be e square square- and serving the lower limit in so B -1 square. So the final answer is hi over to It to about 4 -1 units. Cute.

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