Question

If \[ f(x)=\max\{\sin x,\cos x\} \] and \[ g(x)=\min\{\sin x,\cos x\}, \] then \[ \int_0^{\pi} f(x)\,dx+\int_0^{\pi} g(x)\,dx= \]

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

Hint:

For any two quantities \(a\) and \(b\), always remember that \(\max(a,b)+\min(a,b)=a+b\). This identity avoids unnecessary case-wise integration.

Answer 1

The Correct Option is C

Step 1: Use the relation between maximum and minimum.
For any two real numbers \(a\) and \(b\), \[ \max\{a,b\}+\min\{a,b\}=a+b \] Here, \[ a=\sin x,\qquad b=\cos x \] Therefore, \[ f(x)+g(x)=\sin x+\cos x \]

Step 2: Combine the two integrals.
We have \[ \int_0^{\pi} f(x)\,dx+\int_0^{\pi} g(x)\,dx \] \[ = \int_0^{\pi} \left[f(x)+g(x)\right]\,dx \] \[ = \int_0^{\pi}(\sin x+\cos x)\,dx \]

Step 3: Evaluate the integral.
\[ \int_0^{\pi}(\sin x+\cos x)\,dx = \int_0^{\pi}\sin x\,dx+\int_0^{\pi}\cos x\,dx \] Now, \[ \int_0^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} \] \[ = -\cos\pi+\cos0 \] \[ = -(-1)+1 \] \[ =2 \] Also, \[ \int_0^{\pi}\cos x\,dx = [\sin x]_0^{\pi} \] \[ = \sin\pi-\sin0 \] \[ =0 \] Therefore, \[ \int_0^{\pi}(\sin x+\cos x)\,dx=2+0 \] \[ =2 \]

Step 4: Final conclusion.
Hence, \[ \boxed{2} \]