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

Question

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

Accepted Answer

Solving $x^2/2 = 2x+6 \implies x^2-4x-12=0 \implies (x-6)(x+2)=0$, giving intersection points $(6,18)$ and $(-2,2)$. The second-quadrant point is $(-2,2)$, so $a=-2,\,b=2$.

By King's rule, $I = \displaystyle\int_{-2}^2\frac{9x^2}{1+5^x}dx = \int_{-2}^2\frac{9x^2}{1+5^{-x}}dx$; adding, $2I=\displaystyle\int_{-2}^2 9x^2\,dx$:

$$I = 9\int_0^2x^2\,dx = 9\cdot\frac83 = 24$$

Solutions · 1