MHT CET · Maths · Area Under Curves
The area of the region bounded by the curve \(y=\sin x\) between \(x=-\pi\) and
\(x=\frac{3 \pi}{2}\) is
- A 2 (unit) \(^{2}\)
- B 5 (unit) \(^{2}\)
- C 3 (unit) \(^{2}\)
- D 1 (unit) \(^{2}\)
Answer & Solution
Correct Answer
(B) 5 (unit) \(^{2}\)
Step-by-step Solution
Detailed explanation
(C)
Required area is shaded :
\(\begin{aligned}
A &=2 \int_{0}^{\pi} \sin x d x+\int_{\pi}^{3 \pi / 2} \sin x d x \\
&=2\left[-\left.\cos x\right|_{0} ^{\pi}+[-\cos x]_{\pi}^{\frac{3 \pi}{2}}\right.
\end{aligned}\)
\(=|2[-\cos \pi+\cos 0]|+\left[-\cos \left(\frac{3 \pi}{2}\right)+\cos \pi\right]\)
\(=2[-(-1)+1]+[0+(-1)]\)
\(=2(2)-(1)=5(\text { unit })^{2}\)

Required area is shaded :
\(\begin{aligned}
A &=2 \int_{0}^{\pi} \sin x d x+\int_{\pi}^{3 \pi / 2} \sin x d x \\
&=2\left[-\left.\cos x\right|_{0} ^{\pi}+[-\cos x]_{\pi}^{\frac{3 \pi}{2}}\right.
\end{aligned}\)
\(=|2[-\cos \pi+\cos 0]|+\left[-\cos \left(\frac{3 \pi}{2}\right)+\cos \pi\right]\)
\(=2[-(-1)+1]+[0+(-1)]\)
\(=2(2)-(1)=5(\text { unit })^{2}\)

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