Question

Find area bounded by the curves \(y\) = \(max\{sin x, cos x\}\) and \(x-axis\) between \(x = -  \pi  \)and \(x =   \pi\)

• A. \(2 + \sqrt{2}\)
• B. \(\sqrt{2}\)
• C. \(1 + \sqrt{2}\)
• D. \(2\sqrt{2}\)

Answer 1

The Correct Option is D

Step 1: Determine where \(\sin x\) and \(\cos x\) intersect and which is larger.

We solve \(\sin x = \cos x\) for \(x\in[-\pi,\pi]\). This gives

\(\tan x = 1 \implies x = -\frac{3\pi}{4}\) and \(x=\frac{\pi}{4}\) (within \([-\pi,\pi]\)).

Hence, the interval \([-\pi,\pi]\) is split into three subintervals:

\([-\pi,-\frac{3\pi}{4}], [-\frac{3\pi}{4},\frac{\pi}{4}], [\frac{\pi}{4},\pi]\).

A quick check in each interval shows:

On \([-\pi,-\frac{3\pi}{4}]\), we have \(\sin x > \cos x\), so \(f(x)=\sin x\).

On \([-\frac{3\pi}{4},\frac{\pi}{4}]\), we have \(\cos x \ge \sin x\), so \(f(x)=\cos x\).

On \([\frac{\pi}{4},\pi]\), again \(\sin x \ge \cos x\), so \(f(x)=\sin x\).

Step 2: Find where these maxima are nonnegative (so they contribute positive “area”).

In \([-\pi,-\frac{3\pi}{4}]\), \(\sin x\) is actually negative, so the region above the \(x\)–axis is zero there.

In \([-\frac{3\pi}{4},-\frac{\pi}{2}]\), \(\cos x\) is also negative, so again no area above the axis.

In \([-\frac{\pi}{2},\frac{\pi}{4}]\), \(\cos x\) becomes nonnegative, so the area is \(\int_{-\pi/2}^{\pi/4}\cos x\,dx\).

Finally, in \([\frac{\pi}{4},\pi]\), \(\sin x\) is nonnegative, so the area is \(\int_{\pi/4}^{\pi}\sin x\,dx\).

Step 3: Compute these two integrals and sum.

\(\text{Area}_1 = \int_{-\pi/2}^{\pi/4}\cos x\,dx = [\sin x]_{-\pi/2}^{\pi/4} = \sin(\frac{\pi}{4}) - \sin(-\frac{\pi}{2}) = \frac{\sqrt{2}}{2} - (-1) = 1 + \frac{\sqrt{2}}{2}\).

\(\text{Area}_2 = \int_{\pi/4}^{\pi}\sin x\,dx = [-\cos x]_{\pi/4}^{\pi} = [-\cos(\pi)] - [-\cos(\frac{\pi}{4})] = [-(-1)] - [-\frac{\sqrt{2}}{2}] = 1 + \frac{\sqrt{2}}{2}\).

Hence the total area is

\(\text{Area} = (1 + \frac{\sqrt{2}}{2}) + (1 + \frac{\sqrt{2}}{2}) = 2 + \sqrt{2}\).

But we must not forget the small intervals \([-\pi,-\frac{3\pi}{4}]\) and \([-\frac{3\pi}{4},-\frac{\pi}{2}]\) where the function is negative; the “enclosed area” is the absolute value of that portion. Computing directly (or by symmetry arguments), one finds that adding those negative‐region contributions precisely yields a net of \(+2 - \sqrt{2}\), which combines with the above \(2+\sqrt{2}\) to total 4.

A quicker (and more careful) piecewise‐analysis shows the final sum of absolute areas is exactly:

\(4\).

Thus the region between \(y=f(x)\) and the \(x\)–axis has total area \(\boxed{4}\).