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

Question

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 $\mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{9 x^2}{1+5^x} \mathrm{~d} x$ is equal to :

Accepted Answer

$$\begin{aligned}
& x^2=2 y \text { and } y-2 x-6=0 \
& \frac{x^2}{2}-2 x-6=0 \
& x^2-4 x-12=0 \
& x^2-6 x+2 x-12=0 \
& x(x-6)+2(x-6)=0 \
& (x-6)(x+2)=0
\end{aligned}$$

Point of intersection are $(6,18)$ and $(-2,2)$

$(-2,2)$ is in second quadrant

$$\begin{aligned}
& a=-2, b=2 \
& I=\int_{-2}^2 \frac{9 x^2}{1+5^x} d x\quad\text{..... (i)}
\end{aligned}$$

$I=\int_{-2}^2 \frac{9 x^2}{1+5^{-x}} d x\quad\text{..... (ii)}$

$$\begin{aligned}
&\text { Adding (i) and (ii) }\
&\begin{aligned}
& 2 I=\int_{-2}^2 9 x^2 d x \
& I=9 \int_0^2 x^2 d x \
& I=9\left(\frac{x^3}{3}\right)_0^2 \Rightarrow I=24
\end{aligned}
\end{aligned}$$

Solutions · 1