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  • Need solution for RD sharma maths class 12 chapter 22 area of bounded region exercise 20.2 question 1

Need solution for RD sharma maths class 12 chapter 22 area of bounded region exercise 20.2 question 1

Answers (1)

\frac{7}{3} sq.units

Hint:

               y=4x^{2} is an upward parabola with vertex (0,0)

Given:

               y=4x^{2}

Solution:

x=0 is the line with points (0,1),(0,0) etc.,

y=1 is the line with points (0,1),(1,1) etc.,

y=4 is the line with points (0,4),(1,4) etc.,

Parabola meets at points (0,0) with line x=0

 \left ( \frac{1}{2},1 \right ) with liney=1, (1,4) with line y=4

Here we need to do along y-axis as we don’t know equation of line with respect to x

Required area (Shaded one)=Area under parabola from y=1 to y=4

              \begin{aligned} &\int_{1}^{4}|x| d y=\int_{1}^{4} \sqrt{\frac{y}{4}} d y \\ \end{aligned}

\begin{aligned} &=\frac{1}{2} \int_{1}^{4} \sqrt{y} d y \\ &=\left[\frac{1}{2} \frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4} \\ \end{aligned}

\begin{aligned} &=\frac{1}{2} \times \frac{2}{3}\left[y^{\frac{3}{2}}\right]_{1}^{4} \\ &=\frac{1}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right] \\ \end{aligned}

\begin{aligned} &=\frac{1}{3}(8-1) \\ &=\frac{7}{3} s q \cdot u n \text { its } \end{aligned}  

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