Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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}

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE admit card 2025 out for Class 10, 12 supplementary exams
anam-khan

CBSE Class 10, 12 practical exams 2025 for compartment students from July 10; detailed guidelines issued
anam-khan

CBSE Central Sector Scheme of Scholarship 2025 applications open on scholarships.gov.in

Latest Question