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Step-by-Step Solution
Step 1: Identify the Points of Intersection
The graphs of the functions \sin x and \cos x intersect where
\sin x = \cos x . The standard solutions to this equation are
x = \frac{\pi}{4} + n\pi , for integer n . Two consecutive intersection points can be taken as
x = \frac{\pi}{4} and x = \frac{5\pi}{4} .
Step 2: Determine Which Function Lies Above the Other
Over the interval
\left[\frac{\pi}{4}, \frac{5\pi}{4}\right] , we check sample values (like x = \frac{\pi}{2} ) to see that
\sin x \geq \cos x in this range. Hence, the area enclosed between \sin x and \cos x on this interval is
\displaystyle \int_{\pi/4}^{5\pi/4} (\sin x - \cos x)\, dx .
Step 3: Evaluate the Integral
Compute
A = \int_{\pi/4}^{5\pi/4} \bigl(\sin x - \cos x\bigr)\, dx.
First, find the antiderivative:
\int (\sin x - \cos x)\, dx = -\cos x - \sin x.
Therefore,
A = \left[\, -\cos x - \sin x \,\right]_{\pi/4}^{5\pi/4}.
Substituting the upper limit x = \frac{5\pi}{4} :
-\cos\bigl(\tfrac{5\pi}{4}\bigr) \;-\; \sin\bigl(\tfrac{5\pi}{4}\bigr)
= -\Bigl(-\frac{\sqrt{2}}{2}\Bigr)\;-\;\Bigl(-\frac{\sqrt{2}}{2}\Bigr)
= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}
= \sqrt{2}.
Substituting the lower limit x = \frac{\pi}{4} :
-\cos\bigl(\tfrac{\pi}{4}\bigr) \;-\; \sin\bigl(\tfrac{\pi}{4}\bigr)
= -\frac{\sqrt{2}}{2} \;-\; \frac{\sqrt{2}}{2}
= -\sqrt{2}.
Hence,
A = \sqrt{2} - (-\sqrt{2}) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}.
Step 4: Compute A^4
Since A = 2\sqrt{2} , we get
A^2 = (2\sqrt{2})^2 = 4 \times 2 = 8,
and therefore
A^4 = (A^2)^2 = 8^2 = 64.
Step 5: State the Final Answer
The value of A^4 is
64.
Reference Image for the Solution
