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  • Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20 3 question 15

Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 15.

Answers (1)

Answer:

\frac{27}{2}

Hint:

First we need to find the point of intersection of the given curve

Given: Equation of parabola y=3x^2  and equation of line 3x-y+6=0

Solution:

 Equation of parabola y=3x^2             ---(i)

Equation of line 3x-y+6=0                  ---(ii)

Putting the value of y in equation (ii)

\begin{aligned} &3 x-3 x^{2}+6=0 \\ &-3 x^{2}+3 x+6=0 \\ &3 x^{2}-3 x-6=0 \\ &x^{2}-x-2=0 \\ &x^{2}-2 x+x-2=0 \\ &x(x-2)+1(x-2)=0 \end{aligned}

\begin{array}{ll} (x+1)(x-2) & =0 \\ x+1=0 & , x-2=0 \\ x=-1 & , x=2 \end{array}

When x=-1  then y=3(-1)^2=3

When x=2  then y=3(2)^2=12

So the points of the interaction of the given curves is (1 , -3 ) and (2 , 12)

\text{Required area}= \int_{-1}^{2}(Y_{\text{Line}}-Y_{\text{Parabola}})dx

\begin{aligned} &=\int_{-1}^{2}\left(3 x+6-3 x^{2}\right) d x \\ &=\left[3 \frac{x^{2}}{2}+6 x \frac{-3 x^{2}}{3}\right]_{-1}^{2} \\ &=\left(\frac{3}{2} \times 4+12-8\right)-\left(\frac{3}{2}-6+1\right) \\ &=\frac{10}{1}+\frac{7}{2} \\ &=\frac{20+7}{2} \\ &=\frac{27}{2} \end{aligned}

Hence , area of the region is \frac{27}{2}.

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