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  • Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 7

Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 7

Answers (1)

Answer:

\frac{2}{3}\left(5^{\frac{3}{2}}-1\right)

Hint:

Where y=\sqrt{x+1} \text { in }[0,4]

Given:

Sketch the graph of y=\sqrt{x+1} in \left [ 0,4 \right ] and determine the area of region enclosed by curve, thex-axis and line x=0,x=4

Solution:

 

\text { Area of } O A B C O=\int_{0}^{4}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]

\begin{aligned} &A=\int_{0}^{4} y d x \\\\ &A=\int_{0}^{4} \sqrt{x+1} d x \\\\ &A=\int_{0}^{4}(x+1)^{\frac{1}{2}} d x \end{aligned}

\begin{aligned} &A=\left[\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} \\\\ &A=\frac{2}{3}\left(5^{\frac{3}{2}}-1\right) s q \cdot \text { unit } \end{aligned}                    \left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

 

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