Finding area
Find the area of the region enclosed by the...

Question

Finding area

Find the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:

c. 2π ≤ x ≤ 3π.

Accepted Answer

Welcome back everyone. In this problem we want to find the area of the shaded region Y equals x multiplied by the sin of 2 x. A says it's 3/4 of pi square units, B 1/4 of pi square units, C 5/4 of pi square units, and D 7/4 of pi square units. Now how can we figure out this area for the shaded region? What do we know? Well, recall, OK, that. The area would be equal to the definite integral between the bonds of A and B of F of X with respect to x. So if we can identify the bones and our function, then we should be able to integrate and solve for the area. Now, from our graph we can visually see that the shaded region is from x equals pi to x equals 3 halves of pi. So here a. Or lower bound would be pie. B or upper bound would be three halves of pi andy would be our function of x, which is x multiplied by the sin of 2 x. So this means then. That our area would be equal to the integral between pi and 3 halves of pi. Of xin to x with respect to x. Now how can we integrate our function here? Well, we can use integration by parts. So using integration by parts. Yeah Recall, it basically tells us that our integral of UDV between the bones of A and B would be equal to UV. Limit bound by A and B minus the integral between A and B of VDU. So if we can set one of these terms in our product to you and the other to DV, then we should be able to manipulate them to help us to solve for our integral. So let's set you to be equal to x, OK. And let's say dV to be equal to the sine of 2 x with respect to x. That means then d, the derivative of u with respect to x would be 1, so duu equals dx, and if we integrate both sides for dv, that means v would be equal to the integral of sin 2 x with respect to x, which is negative half of the cosine of 2 x plus c. So now that we have U and V applying the integration by parts formula, this means that our integral, OK. Or integral is going to be equal to. UV, so that's X multiplied by -12 of the cosine of 2 x, OK. Between the bones of pie and 3 halves of pie. Minus Or integral, OK. Between the bones between the same bones of. ZDU, so that's negative half of the cosine of 2 x multiplied by dx. Now if we simplify here for a first term, we could rewrite it as -12 of x multiplied by the cosine of 2 x, OK. Between the bones of. I and the 3 halves of pi, OK, and for a second term we're subtracting negative half of the cosine of 2 x so we could rewrite that as plus a half, OK, of the integral. OK, of class 2 X with respect to X. And now we notice here both of them have the same bones, so we could rewrite them all in the same bound where we're going to have 1/2 x multiplied by the cosine of 2 x, OK, plus a half of this integral of the cosine of 2 x here, which would be a half of sin 2 x, so that's a 12 multiplied by a half or a quarter of the sine of 2 x, OK, between the bones of pi and 3 halves of pi. Now where do we go from here? Well, let's apply our upper and lower bones. So that means our error is going to be -122, OK, of 3 halves of pi multiplied by the cosine of 2 multiplied by 3 halves of pi or the cosine of 3 pi, OK. Plus A quarter of the sign of 3 halves of pi 2 multiplied by 3 halves of pi, sorry, which would also be 3 pi. Minus 12 of pi multiplied by the cosine of 2 multiplied by pi or 2 pi plus a quarter of the sin of 2 pi. I know when we simplify here. In our first term that will be -3/4 of pi, OK, multiplied by -1 + 0. And then our second term, that will be negative or half of pi, OK, multiplied by 1. Plus 0 This simplifies the 3/4 of pie. Plus 1/2 of pi, which would be equal to 5/4 of pi. Therefore, our area is 5/4 of pi square units. When we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Finding areaFind the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:c. 2π ≤ x ≤ 3π.

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Finding area
Find the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:
c. 2π ≤ x ≤ 3π.