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

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(A) $1$. The curve is $y = 2x^2$, which implies $x = \sqrt{y/2} = \frac{\sqrt{y}}{\sqrt{2}}$.$2$. The region is bounded by $x = 1$ (where $y = 2(1)^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) $1$. The curve is $y = 2x^2$, which implies $x = \sqrt{y/2} = \frac{\sqrt{y}}{\sqrt{2}}$.$2$. The region is bounded by $x = 1$ (where $y = 2(1)^2 = 2$) and $y = 4$ (where $x = \sqrt{4/2} = \sqrt{2}$).$3$. The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 2$ to $y = 4$, minus the area of the rectangle formed by $x=1$ and $y=2$ if we integrate w.r.t $y$, or more simply: $A = \int_{1}^{\sqrt{2}} (4 - 2x^2) dx$.$4$. $A = [4x - \frac{2x^3}{3}]_{1}^{\sqrt{2}}$.$5$. $A = (4\sqrt{2} - \frac{2(2\sqrt{2})}{3}) - (4(1) - \frac{2(1)^3}{3})$.$6$. $A = (4\sqrt{2} - \frac{4\sqrt{2}}{3}) - (4 - \frac{2}{3}) = \frac{8\sqrt{2}}{3} - \frac{10}{3} = \frac{8\sqrt{2} - 10}{3}$.

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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