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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $ x $ and $ y $. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

$ y = \sin x $ , $ y = x $ , $ x = \frac{\pi}{2} $ , $ x = \pi $

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Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

Michael M.

May 13, 2021

these are not the right questions that match the book

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

06:16

Sketch the region enclosed…

02:32

05:13

05:14

01:25

02:53

09:51

06:31

02:31

Okay, so we need to draw those given curves. First, let's draw x y axis and then draw the first curve which is 6 okay. So, let's draw sine x, we start from negative pi. This goes to 0 and stop at pi. Then we draw i equals to x, which is something like this: it's like the tangent curve of sine x at the region and x equals to pi over 2 here x equals to pi over 2. This is 1. Sin x equals to 1 point. So here there will be 1 and this is x, equals to pi. Okay, so for the bounded region will be this shaded region right here, so our job is evaluate this area. So how do we do it? We find the approximating, since everything we can see here is represented by x. So when we do the integral we do it with respect to x, okay area, where equals to the integral, with respect to x, all right. So if i so good and we need to find the boundary the boundary for x is just goes from pi over 2 pi, and this here is just upper curve minus the lower curve. Before that we can see, we can draw a typical approximating rectangle, something like this. It'S very small here, if they do me do out, it will look like this is approximating rectangle of the will be delta, x and height will be x. I minus sine x. I okay, so in other words our integral will be x, minus sine x. Okay, then they milorade this integral to find the area of the region. The anti derivative of this is the learned, half x square and sine x. Anti derivative of this is negative cosine. I be plus cosine x, i i plus cosine x. Then we evaluate at x equals pi minus x equal to pi over 2, so the first term there will be pi square over 2 and second, a cosine, pi yo, know cosine. We know cosiniative 1. So this is minus 1 minus x equals to pi over 2, so this will be pi squared over 4 times 1 half is pi square over 8 and cosine pi over 2 is 0, so this plus 0 p, in other words our area. There will be pi square over 2 minus pi square over 8, which is 3 pi square over 8 minus 1 point: okay,

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