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  • Let   be the area of the larger region bounded by the curve y <img alt="2 = 8x" src="https://entran

Let \alpha  be the area of the larger region bounded by the curve y 2 = 8x and the linesy = x  and x = 2, which lies in the first quadrant. Then the value of 3\alpha is equal to

Option: 1

22


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

\operatorname{area}(\alpha)=\int_2^8(2 \sqrt{2} \sqrt{x}-x) d x

\begin{aligned} & =\left[2 \sqrt{2} \cdot \frac{2}{3} \mathrm{x}^{3 / 2}-\frac{x^2}{2}\right]^8 \\ & =\frac{4 \sqrt{2}}{3}[8 \times 2 \sqrt{2}-2 \sqrt{2}]-30 \\ & =\frac{28 \times 4}{3}-30 \\ & =\frac{112}{3}-30 \\ & \alpha=\frac{22}{3} \\ & 3 \alpha=22 \end{aligned}

 

 

 

 

 

 

Posted by

Pankaj Sanodiya

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