Let the curve

Let the curve $y=y(x)$ be the solution of the differential equation, $\frac{\mathrm{dy}}{\mathrm{dx}}=2(\mathrm{x}+1)$. If the numerical value of area bounded by the curve $\mathrm{y}=\mathrm{y}(\mathrm{x})$ and $\mathrm{x}$-axis is $\frac{4 \sqrt{8}}{3}$, then the value of $\mathrm{y}(1)$ is equal to


$\frac{d y}{d x}=2(x+1)$

$\Rightarrow \quad \int d y=\int 2(x+1) d x$

$\Rightarrow y(x)=x^{2}+2 x+C$

Area $=\frac{4 \sqrt{8}}{3}$


$\Rightarrow \quad 2 \int_{-1}^{-1+\sqrt{1-C}}\left(-(x+1)^{2}-C+1\right) d x=\frac{4 \sqrt{8}}{3}$

$\Rightarrow \quad 2\left[-\frac{(x+1)^{3}}{3}-C x+x\right]_{-1}^{-1+\sqrt{1-C}}=\frac{4 \sqrt{8}}{3}$

$\Rightarrow \quad-(\sqrt{1-C})^{3}+3 c-3 C \sqrt{1-C}$

$-3+3 \sqrt{1-\mathrm{C}}-3 \mathrm{C}+3=2 \sqrt{8}$

$\Rightarrow \quad C=-1$

$\Rightarrow \quad f(x)=x^{2}+2 x-1, \quad f(1)=2$



Leave a comment

Please enter comment.
Please enter your name.