Area: Find the area of the region bounded above by y = 2 cos x...

Question

Area: Find the area of the region bounded above by y = 2 cos x and below by y = sec x, −π/4 ≤ x ≤ π/4.

Accepted Answer

Welcome back everyone. Let R be the region in the plane bounded above by y equals 3 sin x and below by y equals tangent x over the interval from 0 up to pi divided by 4 inclusive. Compute the exact area of r. So for this problem we want to identify the area a. We can write subscript r. And we have to recall that it is a definite integral in this case from 0 up to pi divided by 4, and we want to integrate the difference. Let's remember that we take the top function, it says above. By y equals 3 sin x, so the top function is 3 sin x, and we're going to subtract the bottom function. It says below by y equals tangent x, so we're subtracting tangent x. We're integrating with respect to x, and we can now start integrating. Starting with 3 sin x, the integral of sin is negative cosine, so we get -3 cosine x, and the integral of tangent x is ln of the absolute value of sequent x. So we're going to subtract ln of the absolute value of sequent x, and this is how we get the result. What we want to do is simply evaluate. From 0 to pi divided by 4 using the fundamental theorem of calculus part 2. So now let's substitute. The limits we're going to get -3 cosine of pi divided by 4 minus ln of the absolute value of. We can rewrites as 1 divided by cosine, so that would be 1 divided by cosine of pi divided by 4. And then we are subtracting the value of the function at 0, so minus. In parentheses we are going to have 3 cosine 0 minus ln of the absolute value of 1 divided by cosine 0. From here, our final step is to. Substitute the values, so we're going to have -3 multiplied by the square root of 2 divided by 2. That's cosine of pi divided by 4 minus ln of 1 divided by the square root of 2 divided by 2. We can just flip the denominator, so that'll be ln of square root of 2. Because 2 divided by the square root of 2 gives us the square root of 2. And now if we rewrite the square root of 2 as 2 raise the power of 1/2, we can apply the properties of logarithms and we can Bring down the exponent. Then we are going to add, because we have negative negative, right, 3 cosine 0, so that's just 3 because cosine of 0 is 1, and ln of 1 is 0, so we don't need to add the final term. Now the final answer is going to be 3 minus 3 square root of 2. Divided by 2. Minus 1/2 ln of 2, right? And we are going to add the units, those will be square units. Let's label the final answer, and thank you for watching.

Area: Find the area of the region bounded above by y = 2 cos x and below by y = sec x, −π/4 ≤ x ≤ π/4.

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