Let $(a, b)$ be the point of intersection of the curve $x^2=2 y$ and the straight line $y-2 x-6=0$ in the second quadrant. Then the integral $I=\int_a^b \frac{9 x^2}{1+5^x} d x$ is equal to :

Question

Aicharya · Accepted Answer

$\begin{aligned} & x^2=2 y \& y=2 x+6 \ & x^2=4 x+12\end{aligned}$ $\left.x^2-4 x-12=0 \Rightarrow \begin{gathered}x=6 \ y=18\end{gathered} ightvert\, \begin{gathered}	ext { if } x=-2 \ y=2\end{gathered}$ $	herefore(6,18) \&(-2,2)$ Here $(6,18)$ Rejected because $(a, b)$ lies in $2^{	ext {nd }}$ quadrant $\begin{aligned} & 	herefore \mathrm{a}=-2 \& \mathrm{~b}=2 \ & 	herefore \mathrm{I}=\int_{-2}^2 \frac{9 \mathrm{x}^2}{1+5^{\mathrm{x}}} \mathrm{dx}=\int_{-2}^2 \frac{9 \cdot 5^{\mathrm{x}} \cdot \mathrm{x}^2}{1+5^{\mathrm{x}}} \mathrm{dx} \ & 	herefore 2 \mathrm{I}=\int_{-2}^2 9 \mathrm{x}^2 \mathrm{dx}=18 \int_0^2 \mathrm{x}^2 \mathrm{dx}=18\left(\frac{\mathrm{x}^3}{3}ight)_0^2 \ & 2 \mathrm{I}=48 \ & 	herefore \mathrm{I}=24\end{aligned}$

Let $(a, b)$ be the point of intersection of the curve $x^2=2 y$ and the straight line $y-2 x-6=0$ in the…

Solution