Find the area of the region enclosed by the curves y = √(2)cos x and y = (1)/(2)sec^2 x, -(π)/(4

Step 1: Find the points of intersection of the two curves y = sqrt(2)cos(x) and y = (1/2)sec^2(x

Find the area of the region enclosed by the curves y =...

Find the area of the region enclosed by the curves $y = \sqrt{2} \cos x$ and $y = \frac{1}{2} \sec^2 x$, $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$. The area is $oxed{ }$. (Type an exact answer, using radicals as needed.)

Transcript

00:01 To help us with our calculations, we'll call 4 cosine x, f of x, and 4 cosine 2x, g of x.

00:07 And so to find the area of the shaded region, we'll use the fact that the area of a region from a to b is the integral of the upper function minus the lower function, where the upper function is greater than the lower function.

00:30 So notice that if we call this point k, then in the interval 0 to k, we have that f of x is greater than g of x, and that from k to pi, g of x is greater than f of x.

00:43 So the area of the region is the integral from 0 to k of f of x of x minus g of x plus the integral from k to pi of g of x minus f of x.

00:59 So we need to find the value of k.

01:01 So we can do this by equating f of x and g of x.

01:03 So we get 4 cosine x equals to 4 cosine 2x.

01:10 So we can cancel out the fours.

01:13 And cosine 2x can be rewritten as 2 cosine square x minus 1.

01:19 We can subtract cosine x from both sides to get 2 cosine square x minus cosine x minus 1 equals 0.

01:26 So that's a quadratic equation.

01:28 So to factor this equation, we can write 2 cosine square x minus 2, cosine x plus cosine x minus 1 so on the first two terms we can factor out to cosine x from cosine x minus 1 we can factor out 1 from cosine x minus 1 now we can factor out the cosine x minus 1 from 2 cosine x plus 1 and now we have that cosine x minus 1 equals to 0 or 2 cosine x plus 1 equals to 0.

02:15 So we have that cosine x equals to 1, and cosine x equals to minus half.

02:23 So in the interval, 0 to pi, cosine x equals to 1 tells us that x equals to 0, and cosine x minus half tells us that x equals to 2 pi over 3.

02:38 So looking back at the graph, we see that k equals to 2 pi over 3.

02:48 So we can rewrite this integral with interval 0 to 2 pi over 3 of f of x minus g of x plus the integral from 2 pi over 3 to pi of x.

03:06 So to help make our calculation simpler, we can rewrite this expression.

03:10 First integral, we can write the same.

03:12 The second integral we can subtract it from the first integral and have f of x minus g of x as our integrand so that the terms are the same in both integrals and we can change the bounds of the second integral so we can switch the bounds around to get plus two pi over three as the upper bound and pi is the lower bound and the integral is f of x minus g of x so what we can do is evaluate the integral from a to b of f of x minus g of x, and plug in the values for the bounds for each of the integrals that we have for the expression corresponding to the area of the region...

Question

Find the area of the region enclosed by the curves $y = \sqrt{2} \cos x$ and $y = \frac{1}{2} \sec^2 x$, $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$. The area is $oxed{ }$. (Type an exact answer, using radicals as needed.)

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Find the area of the region enclosed by the curves $y = \sqrt{2} \cos x$ and $y = \frac{1}{2} \sec^2 x$, $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$. The area is $oxed{ }$. (Type an exact answer, using radicals as needed.)