Question

Find area of region enclosed by the parabola x2=y and the line 4x-y+12=0

Answer 1

Answer:

$\frac{256}{3}$ is the required area.

Step-by-step explanation:

We have been given the parabola $x^2=y$ and line $4x-y+12=0$

We need to find the area of region enclosed by parabola minus line.

$\int\limits^{6}_{-2} ({4x+12-x^2})\, dx$

$\int\limits^{6}_{-2} {4x} \, dx+\int\limits^{6}_{-2} {12} \, dx+\int\limits^{6}_{-2} {-x^2} \, dx$

 $\left | \left ( 2x^2+12x-\frac{x^3}{3} \right ) \right |_{-2}^{6}\\\ \left (2(6)^2+12(6)-\frac{(6)^3}{3} \right )-\left(2(-2)^2+12(-2)-\frac{(-2)^3}{3} \right )\\\\(72+72-72)-(8-24+\frac{8}{3})\\\\\frac{256}{3}$