Q.20 The value of the integral
0
| cos |x dx
is _____
Understanding the integral \(\int_{0}^{\pi}|\cos x|dx\)
On the interval \([0,\pi]\), the function \(\cos x\) changes sign at \(x=\pi/2\).
- For \(0 \le x \le \pi/2\), \(\cos x \ge 0\) so \(|\cos x| = \cos x\).
- For \(\pi/2 \le x \le \pi\), \(\cos x \le 0\) so \(|\cos x| = -\cos x\).
Therefore, the integral is split at \(\pi/2\):
\[
\int_{0}^{\pi}|\cos x|dx
= \int_{0}^{\pi/2}\cos x\,dx
+ \int_{\pi/2}^{\pi}(-\cos x)\,dx.
\]
Step-by-step solution
First part
Evaluate the first integral:
\[
\int_{0}^{\pi/2}\cos x\,dx
= \left[\sin x\right]_{0}^{\pi/2}
= \sin\left(\frac{\pi}{2}\right) – \sin 0
= 1 – 0 = 1.
\]
Second part
Evaluate the second integral:
\[
\int_{\pi/2}^{\pi}(-\cos x)\,dx
= \left[-\sin x\right]_{\pi/2}^{\pi}
= -\sin\pi – \left(-\sin\frac{\pi}{2}\right)
= 0 – (-1) = 1.
\]
Total value
Add the two results:
\[
\int_{0}^{\pi}|\cos x|dx = 1 + 1 = 2.
\]
Geometrically, this integral equals the total area under the curve \(y = |\cos x|\) from \(x = 0\) to \(x = \pi\), which is 2 square units.
Typical MCQ options and explanation
| Option | Value | Explanation |
|---|---|---|
| (A) | 0 | This would be the value of \(\int_{0}^{\pi}\cos x\,dx\) without the absolute value, where positive and negative areas cancel each other; here the absolute value keeps both contributions positive, so this option is incorrect. |
| (B) | 1 | This equals only one of the two partial integrals and ignores the second positive area, so it is not the value of the full definite integral. |
| (C) | 2 | This is the sum of the two positive areas after splitting at \(\pi/2\), so it is the correct value of \(\int_{0}^{\pi}|\cos x|dx\). |
| (D) | \(\pi\) | This might be confused with integrals like \(\int_{0}^{\pi}1\,dx = \pi\); it does not arise from the actual calculation here, so this option is incorrect. |
