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Make the given changes in the indicated examples of this section and then solve the resulting problems.In Example 3 , change $x=4$ to $x=2$ and compare the results with those of Exercise 1.

Calculus 1 / AB

Chapter 25

Integration

Section 3

The Area Under a Curve

Integrals

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

00:41

From here on, each exercis…

01:16

These exercises correspond…

01:25

Solve the equations (In th…

01:03

Watch the section lecture …

03:17

Solve each equation by usi…

00:32

04:23

In Exercises $1-4$ , ident…

01:34

Solve each problem. See Ex…

00:58

Describe similarities and …

01:42

If the exercise is an equa…

Okay, so we're looking at the function f of X equals x squared plus one. And we're gonna be integrating from X is equal to +02 X is equal to two. So we're going to find the area under the curve from X is equal to zero, checks equal to two. So if we set up the integral, go from 0 to 2 for a function which is X squared plus one and we have the dx. And so to find an integral like this, we're just going to use um anti differentiation techniques involving the power rule. So if you remember the derivative of X races, some power end is just n times X rays, the n minus one power. So if we're trying to find the integral, we took the integral. Both sides. The integral of a derivative of something is just going to be what that something is. So you'd have X to the n is equal to the integral of n times X, the n minus one. And so this is what we're gonna do use we're gonna use the fact that we have the integral of X. The n minus one is equal to X. The N. Um Where in this case we're gonna have X squared. So we're gonna get if if X squared is equal to X the n minus one, the next the end is X to the third. And if N is equal to three then we need to um actually put a one third here so that we don't get a three in our derivative since we don't actually have three X squared here, we just have X squared. So this is the technique that we're gonna use to find the integral here. So this would be equal to X. The 3rd divided by three plus X. And we can see if we took the derivative of X the third divided by three. We would bring the three down, three divided by three is one. And then we'll be left with X squared. And if we took the derivative of X it would just be one. So would be left with our original function F. X. So and then we also have the plus C. See in this case isn't going to matter were you were doing a definite integral and we don't have any um parameters where C. Is going to matter so we can ignore see for now. So we're going to look at extra 3rd Divided by three plus X From 0 to 2. So when X is equal to zero, just have zero to third divided by three plus zero which is zero. So we really just need to look at when X is equal to 22 to the third is eight. So we have eight thirds plus two which is equal to Two is equal to 6/3. So this is equal to 14/3. And now if we look at the values, we got an exercise one We got two values, one where N is equal to to where we were looking at. Um two rectangles are area, there was equal to three and the other one was when it was equal to four. In our area there was equal to 15/4. So we can see that the area that we found in exercise one was smaller than the actual area. Um We got when we integrated um the function.

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