sketch the region enclosed by the given curves. decide whether to integrate with respect to x or y. then…

Explanation:

Step1: Determine the upper and lower functions

For (x\in[-\frac{\pi}{4},\frac{\pi}{4}]), we know that (7\cos x\geq81\sec^{2}x). The formula for the area between two curves (y = f(x)) and (y = g(x)) from (x = a) to (x = b) is (A=\int_{a}^{b}[f(x)-g(x)]dx), where (f(x)=7\cos x) and (g(x)=81\sec^{2}x), (a =-\frac{\pi}{4}), (b=\frac{\pi}{4}).

Step2: Use the integral formula for area

[ \begin{align*} A&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(7\cos x - 81\sec^{2}x)dx\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}7\cos xdx-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}81\sec^{2}x dx \end{align*} ] Since (y = 7\cos x) and (y = 81\sec^{2}x) are even functions ((f(-x)=f(x))), we can use the property (\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx) for even functions. [ \begin{align*} A&=2\left(\int_{0}^{\frac{\pi}{4}}7\cos xdx-\int_{0}^{\frac{\pi}{4}}81\sec^{2}x dx\right)\ &=2\left[7\sin x|{0}^{\frac{\pi}{4}}-81\tan x|{0}^{\frac{\pi}{4}}\right] \end{align*} ]

Step3: Evaluate the definite integrals

For (\int_{0}^{\frac{\pi}{4}}7\cos xdx), using the antiderivative (\int\cos xdx=\sin x + C), we have (7\sin x|{0}^{\frac{\pi}{4}}=7\left(\sin\frac{\pi}{4}-\sin0\right)=7\times\frac{\sqrt{2}}{2}-0=\frac{7\sqrt{2}}{2}) For (\int{0}^{\frac{\pi}{4}}81\sec^{2}x dx), using the antiderivative (\int\sec^{2}x dx=\tan x + C), we have (81\tan x|_{0}^{\frac{\pi}{4}}=81(\tan\frac{\pi}{4}-\tan0)=81(1 - 0)=81)

[ \begin{align*} A&=2\left(\frac{7\sqrt{2}}{2}-81\right)\ &=7\sqrt{2}-162 \end{align*} ]

Answer:

(7\sqrt{2}-162)