Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 2

Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 2

Answers (1)

best_answer

Answer: \frac{17}{2}sq\cdot units

Hint: Use   - x+ y = 1

Given: Using integration, find area of region bounded by line y-1=x the x-axis and ordinatex=-2 and x=3

Solution:

Here y-1=x is equation of line

We can write

\begin{aligned} &y=x+1 \\\\ &-x+y=1 \end{aligned}

We get \frac{-x}{1}+\frac{y}{1}=1

Consider AB as line intersecting the x-axis at point C(-1,0)

So required area=\text { Area of } C D A C+\text { Area of } C B E C

=\int_{-1}^{3} y d x+\int_{-2}^{-1}-(y) d x

Substituting the value of y,

\begin{aligned} &=\int_{-1}^{3}(x+1) d x+\int_{-2}^{-1}-(x+1) d x \\\\ &=\left[\frac{x^{2}}{2}+x\right]_{-1}^{3}-\left[\frac{x^{2}}{2}+x\right]_{-2}^{-1} \end{aligned}                   \left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

Substituting the value of x ,

\begin{aligned} &=\left(\frac{9}{2}+3\right)-\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}-1\right)-(2-2) \\\\ &=\left(\frac{15}{2}+\frac{1}{2}\right)-\left(\frac{-1}{2}\right) \\\\ &=8+\frac{1}{2} \\ &=\frac{17}{2} s q \cdot \text { units } \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 Board exams twice a year from 2025, no plan for semesters: MoE
anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question