MHT CET · Maths · Area Under Curves
The area bonded by the curve \(y=\sin ^{2} x, x\) -axis and the lines \(x=0\) and \(x=\frac{\pi}{2}\) is
- A 1 sq. units
- B \(\frac{\pi}{8}\) sq. units
- C \(\frac{\pi}{4}\) sq. units
- D \(\frac{\pi}{2}\) sq. units
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\) sq. units
Step-by-step Solution
Detailed explanation
Required area is shaded.
\(
\begin{aligned}
A &=\int_{0}^{\pi / 2} \sin ^{2} x d x \\
&=\int_{0}^{\pi / 2} \frac{1-\cos 2 x}{2} d x \\
&=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}}-\frac{1}{4}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\
&=\frac{\pi}{4}-0=\frac{\pi}{4}
\end{aligned}
\)
\(
\begin{aligned}
A &=\int_{0}^{\pi / 2} \sin ^{2} x d x \\
&=\int_{0}^{\pi / 2} \frac{1-\cos 2 x}{2} d x \\
&=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}}-\frac{1}{4}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\
&=\frac{\pi}{4}-0=\frac{\pi}{4}
\end{aligned}
\)
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