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:

b. π ≤ x ≤ 2π.

Graph of y = x sin(x) from 0 to 3π showing curve crossing x-axis at π and 2π with labeled axes.

Accepted Answer

Welcome back everyone. In this problem we want to find the area of the shaded region for Y equals x multiplied by the sin of 2 X. A says it's 1/4 of pi square units, B 1/2 of pi square units, C, 3/4 of pi square units, and D, 1/4 square units. Now what do we know that can help us to find the area of this region? Well, recall that the area would be equal to the definite integral between the bounds of a and b of our function f of x with respect to x. So if we can identify our bones and our functions, 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 0 to x equals 1/2 of pi. So X is on the interval from 0 to 1/2 of pi, which means, OK. Which means, or rather let me just say here, which means that A would be equal to 0 and B would be equal to 12 of pi. On the other hand, our function is Y. OK, so Y equals F of X, which is x multiplied by the sin of 2 X. So for this shaded region, it's area then. Would be equal to the definite integral between 0 and 12 of pi of x sin of 2 x with respect to x. So let's go ahead and integrate and then solve. Now, to evaluate this integral, we can use integration by parts. So using integration by parts, OK. To solve for our indefinite the indefinite integral, OK, or for the definite integral that is, we know that it basically tells us our integral between the limits of A and B of UDV would be equal to UV. Between the bones of A and B minus or definite integral between A and B of VDU. So if we can choose U and DV from our product here X sign off to X, then we should be able to use integration by parts, we should be able to evaluate. Now, here. If we let u equal x, then the derivative of u would be equal to dx. DU equals dx because the derivative of u with respect to x would be 1. On the other hand, if we let dv be equal to the sine of 2 x with respect to x, if we integrate both sides, then v would be the integral of sin 2 x, which would be equal to negative half of the cosine of 2 x plus some constant c. So now, if we apply the integration by parts formula, this means our integral of x sin of 2 x with respect to x. Would be equal to UV, so that's x multiplied by a negative half of cost of 2 x, OK. Between the bones of well. Let me write that properly here, yes, between the bones of 01/2 of pi, OK, minus or integral between 0 and 1/2 of pi of VDU. So that's negative half of the cosine of 2 x dx. No, when we go ahead and simplify here, OK, then we'll have. Here, well, we, we need to, we already know what this part is, OK, but no, in this case. We can rewrite it at the first term as negative half of x multiplied by the cosine of 2 x, OK. Between 0 and 1/2 of pi. OK. Minus Yeah, minus or plus rather. Because if we're subtracting a negative, we can rewrite that as plus and take out our constant, so that's plus a half of the integral, OK. Of the cosine of to x with respect to x. We know that integral will be a half of sin 2 x, OK, so then this is going to be a negative 12 x cosine of 2 X. OK. Plus a quarter of the sine of twix, OK, which would also be between the bones. OK, of 0.5 of pi, so both of them are really between these bones, which means then that we can just rewrite it simply all together, OK, so -12 x + 2 x plus a quarter of the sin of 2 x all between the bones, OK. Of 0.5 of pi. So now let's apply our bones. So for our lower bound, this is going to be, or sorry, for our upper bound, that will be -12 multiplied by 1/2 of pi multiplied by the cosine of 2 multiplied by 1/2 of pi, OK. Which is going to be I. Plus A quarter of the sin of 2 multiplied by 1/2 of pi or the sin of pi, OK. So, and I think here I need to take out one of these parentheses. OK, so this is for upper bone, and likewise for a lower bound it's going to be -5 multiplied by 0 of the cosine of 2 multiplied by 0 or 0 plus 25 of the sin of 2 multiplied by 0 or 0. Now when we value it for our first term, this will be 25 multiplied by -1, which is 25 of pi multiplied by -1, which leaves us with a quarter of pi, and for a second term, all of this will simplify to zero. So our area then is a quarter of pi square units. If we look back at our answer choices, that means A 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:b. π ≤ x ≤ 2π.

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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:
b. π ≤ x ≤ 2π.