Mathematics - Area Under Curves Question with Solution | TestHub

The region is bounded by the curve $y=2x^2$, the vertical line $x=1$, and the horizontal line $y=4$.

The intersection of $y=2x^2$ and $y=4$ in the region $x\ge 1$ is given by:

$2x^2=4\Rightarrow x^2=2\Rightarrow x=\sqrt{2}$

The required area $A$ is the integral of the upper boundary minus the lower boundary from $x=1$ to $x=\sqrt{2}$:

$A={\int }_1^{\sqrt{2}}(4-2x^2)dx$

$A={\left[4x-\frac{2x^3}{3}\right]}_1^{\sqrt{2}}$

$A=\left(4\sqrt{2}-\frac{4\sqrt{2}}{3}\right)-\left(4-\frac{2}{3}\right)$

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

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