Question

Find the area of the region enclosed by the parabola x^2=y, the line 4x-y+12=0 and the x axis

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.

Required area is lower figure minus upper figure.

$\int\limits^{6}_{-2} ({4x+12-x^2})\, dx\int\limits^{6}_{-2} {4x} \, dx+\int\limits^{6}_{-2} {12} \, dx+\int\limits^{6}_{-2} {-x^2} \, dx\left | \left ( 2x^2+12x-\frac{x^3}{3} \right ) \right |_{-2}^{6}\\\ \left (2(6)^2+12(6)-\frac{(6)^3}{3} \right )-\left(2(-2)^2+12(-2)-\frac{(-2)^3}{3} \right )\\\\(72+72-72)-(8-24+\frac{8}{3})\\\\\frac{256}{3}$

Answer 2

Answer:

The area is 90.67 square units.

Step-by-step explanation:

Given the parabola $x^2=y$, the line $4x-y+12=0$. we have to find the area of region enclosed by these two and x-axis.

$f(x)=4x-y+12=0$

$g(x)=x^2-y=0$

To find the limits we have to solve the two curves in order to find the intersection point.

$4x-y+12=0\\\\4x-x^2+12=0\\\\x^2-4x-12=0\\\\x^2-6x+2x-12=0\\\\x(x-6)+2(x-6)=0\\\\(x+2)(x-6)=0$

we get the intersection points (-2,4) and (6,36)

Hence, required area i.e the area of the region enclosed between parabola and straight line is

$A=\int\limits^6_{-2} [{f(x)-g(x)}] \, dx$

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

         =$|\frac{4x^2}{2}+12x-\frac{x^3}{3}|$between the limits -2 to 6

         =$72+72-72-8+24+\frac{8}{3}$

         =90.67 square units.