The area enclosed by the curve y = 2x^2 and t | vedclass

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Question

(A) Step $1$: Identify the intersection points. The curve is $y = 2x^2$. Given lines are $x = 1$ and $y = 4$. At $x = 1$, $y = 2(1)^2 = 2$. At $y = 4$

Vedclass · Accepted Answer

$\frac{8\sqrt{2} - 10}{3}$

Vedclass · Answer

(A) Step $1$: Identify the intersection points. The curve is $y = 2x^2$. Given lines are $x = 1$ and $y = 4$. At $x = 1$, $y = 2(1)^2 = 2$. At $y = 4$, $2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2}$.Step $2$: The area is bounded by $x=1$ to $x=\sqrt{2}$ under the line $y=4$ minus the area under the curve $y=2x^2$.Step $3$: Area $A = \int_{1}^{\sqrt{2}} (4 - 2x^2) \, dx$.Step $4$: Integrate: $A = [4x - \frac{2x^3}{3}]_{1}^{\sqrt{2}}$.Step $5$: Substitute limits: $A = (4\sqrt{2} - \frac{2(\sqrt{2})^3}{3}) - (4(1) - \frac{2(1)^3}{3}) = (4\sqrt{2} - \frac{4\sqrt{2}}{3}) - (4 - \frac{2}{3}) = \frac{8\sqrt{2}}{3} - \frac{10}{3} = \frac{8\sqrt{2} - 10}{3}$.

The area enclosed by the curve $y = 2x^2$ and the lines $x = 1$ and $y = 4$ in the first quadrant is ..... sq. units.

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