MHT CET · Maths · Area Under Curves
Area (in sq.units) lying in the first quadrant and bounded by the circle \(x^2+y^2=4\) and the lines \(x=0\) and \(x=2\) is
- A \(\pi\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{3}\).
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(A) \(\pi\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & x^2+y^2=4 \\ & \Rightarrow y^2=4-x^2\end{aligned}\)
\(\begin{aligned} & =\int_0^2 \sqrt{4-x^2} d x \\ & =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\cdot \frac{x}{2}\right)\right]_0^2\end{aligned}\)
\(=\left[\frac{2}{2} \sqrt{4-(2)^2}+\frac{4}{2} \sin ^{-1}\left(\frac{2}{2}\right)-\frac{0}{2} \sqrt{4-0}-\frac{4}{2} \sin ^{-1}\left(\frac{0}{2}\right)\right]\)
\(\begin{aligned} & =2 \times \frac{\pi}{2} \\ & =\pi \text { sq. }\end{aligned}\)
\(\begin{aligned} & =\int_0^2 \sqrt{4-x^2} d x \\ & =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\cdot \frac{x}{2}\right)\right]_0^2\end{aligned}\)
\(=\left[\frac{2}{2} \sqrt{4-(2)^2}+\frac{4}{2} \sin ^{-1}\left(\frac{2}{2}\right)-\frac{0}{2} \sqrt{4-0}-\frac{4}{2} \sin ^{-1}\left(\frac{0}{2}\right)\right]\)
\(\begin{aligned} & =2 \times \frac{\pi}{2} \\ & =\pi \text { sq. }\end{aligned}\)
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