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  • Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 23

Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 23

Answers (1)

Answer:

4\; s q \cdot { unit }

Hint:

Curve, y=\cos x

Given:

Find area by curve y=\cos x and ordinate x=0 and x=2\pi

Solution:

\begin{aligned} &A=\int_{0}^{2 \pi}|y| d x \\\\ &A=\int_{0}^{\frac{\pi}{2}}|y| d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}|y| d x+\int_{\frac{3 \pi}{2}}^{2 \pi}|y| d x \end{aligned}

\begin{aligned} &A=\int_{0}^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}-y d x+\int_{\frac{3 \pi}{2}}^{2 \pi} y d x \\\\ &A=\int_{0}^{\frac{\pi}{2}} \cos x d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}-\cos x d x+\int_{\frac{3 \pi}{2}}^{2 \pi} \cos x d x \end{aligned}

\begin{aligned} &A=[\sin x]_{0}^{\frac{\pi}{2}}+[-\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}+[\sin x]_{ \frac{3 \pi}{2}}^{2\pi} \\\\ &A=1+[1+1]+0-(-1) \\\\ &A=4 \; \mathrm{sq} \cdot \mathrm{unit} \end{aligned}

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