Question

Area of the region bounded by the curve \( y = \cos x \) between \( x = -\frac{\pi}{2} \) and \( x = \pi \) is ----------------

• A. 3 sq units
• B. 4 sq units
• C. 1 sq units
• D. 2 sq units

Hint:

When finding area bounded by a curve, always check whether the curve lies above or below the \(x\)-axisUse modulus for the part below the \(x\)-axis.

Answer 1

The Correct Option is D

Step 1: Write the required area.
Since area is always positive, we take modulus where the curve is below the \(x\)-axis.
\[ A = \int_{-\frac{\pi}{2}}^{\pi} |\cos x|dx \]

Step 2: Identify sign of \( \cos x \).
\[ \cos x \geq 0 \text{ on } \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \]
and
\[ \cos x \leq 0 \text{ on } \left[\frac{\pi}{2},\pi\right] \]

Step 3: Split the integral.
\[ A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\,dx - \int_{\frac{\pi}{2}}^{\pi} \cos x\,dx \]

Step 4: Integrate.
\[ \int \cos x\,dx = \sin x \]

Step 5: Evaluate first integral.
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\,dx = \sin\frac{\pi}{2}-\sin\left(-\frac{\pi}{2}\right) \]
\[ = 1-(-1)=2 \]

Step 6: Evaluate second integral.
\[ -\int_{\frac{\pi}{2}}^{\pi} \cos x\,dx = -\left[\sin x\right]_{\frac{\pi}{2}}^{\pi} \]
\[ =-(0-1)=1 \]

Step 7: Final conclusion.
\[ A = 2+1=3 \]
\[ \boxed{3 \text{ sq units}} \]