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Sketch the region enclosed by the given curves an…

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Problem 15 Easy Difficulty

Sketch the region enclosed by the given curves and find its area.

$ y = \sec^2 x $ , $ y = 8 \cos x $ , $ \frac{-\pi}{3} \le x \le \frac{\pi}{3} $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

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Video Transcript

On this problem, we need to sketch the reeds been enclosed by those 2 curves. So let's draw this x y plane first and as we can see, the second 1 is much easier to draw and we know our x. Our domain goes from negative pi over 3 to pi over 3, so assuming this is pi over 3. This is negative pi over 3 and here is positive, pi over 3 or cosine x. The goes to 0 point. We know this is a given function. Sorry, symmetric along y exists, so x, equal to 0 is equals to 8 and we know that x equals to pi over 3 cosine pi over 3 is half so this equals to 4 since its symmetric. This also equals 4 point. So y equals to 8 cosine x should look like this okay. This is some y equals to cosine, 8 cosine x and the second. The first term here is second x square, and we know the second x square is just over. Cosine x square is 1 over cold, fine, x square, and so the x equals to 0. This equals 1 and x equals 2 pi over 3. On this cosine power, 3 is put half so 1 half square is 1 quarter and the reciprocal of our quarter is 4. Actually, intersection point is exactly of the ends of cosine x to the end poinsnegative pi over 3 and a positive pi over 3. These functions should look like this. We got this enclosed area now. Our next step is to find this area. So, as we see, all the things represented by x will take the integral with respect to x and the boundary we already found is just a negative pi over 3 to pi over 3 and the thing inside the integral is upper curve. Upper curve is 8 cos. Minus the lower curve, which is a second x square now, let's find the antidotive of this, so the first term will be 8 sine x. They know the antidote of cosine x is just 6. If you remember, anti derivative for second x is actually tangent x. If you don't remember, you can check the charge of the anti derivative, then we evaluate the boundary x equals to pi over 3 and x, equals to negative pi over 3, and then our x equals to pi over 3. We know sine pi over 3 square root. 3 over 2 point, so these 4 square root, 3 minus and tangent tangent pi over 3 square root, 3 minus the plug and pi x equals to the negative power 3. So this will be negative. 4 square 3 plus square root 3 right. So our answer will be 6 square root. Thre.

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