for each problem, find the area of the region enclosed by the curves. you may use the provided graph to…

for each problem, find the area of the region enclosed by the curves. you may use the provided graph to sketch the curves and shade the enclosed region.\n\n8) $y = -2 \\cdot \\sec^2 x$\n$y = 2 \\cos x$\n$x = 0$\n$x = \\frac{\\pi}{4}$

for each problem, find the area of the region enclosed by the curves. you may use the provided graph to sketch the curves and shade the enclosed region.\n\n8) $y = -2 \\cdot \\sec^2 x$\n$y = 2 \\cos x$\n$x = 0$\n$x = \\frac{\\pi}{4}$

Answer

Explanation:

Step1: Identify the upper and lower functions

On the interval $[0, \frac{\pi}{4}]$, $2\cos x \geq 0$ and $-2\sec^2 x \leq -2$. Thus, $y_{upper} = 2\cos x$ and $y_{lower} = -2\sec^2 x$.

Step2: Set up the definite integral for area

$$A = \int_{0}^{\frac{\pi}{4}} (2\cos x - (-2\sec^2 x)) , dx$$

Step3: Simplify the integrand

$$A = \int_{0}^{\frac{\pi}{4}} (2\cos x + 2\sec^2 x) , dx$$

Step4: Find the antiderivative

$$\int (2\cos x + 2\sec^2 x) , dx = 2\sin x + 2\tan x$$

Step5: Evaluate at the boundaries

$$A = [2\sin x + 2\tan x]_{0}^{\frac{\pi}{4}}$$

Step6: Calculate the final value

$$A = (2\sin \frac{\pi}{4} + 2\tan \frac{\pi}{4}) - (2\sin 0 + 2\tan 0) = 2(\frac{\sqrt{2}}{2}) + 2(1) - 0$$

Step7: Simplify the result

$$A = \sqrt{2} + 2$$

Answer:

$\sqrt{2} + 2$