The area (in sq. units) of the region {(x, y) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The region is bounded by the parabola $y = 4x^2$ and the line $y = 8x + 12$.To find the points of intersection,set $4x^2 = 8x + 12$.$4x^2 - 8x - 1

Vedclass · Accepted Answer

$\frac{128}{3}$

Vedclass · Answer

(D) The region is bounded by the parabola $y = 4x^2$ and the line $y = 8x + 12$.To find the points of intersection,set $4x^2 = 8x + 12$.$4x^2 - 8x - 12 = 0 \implies x^2 - 2x - 3 = 0$.$(x - 3)(x + 1) = 0$,so $x = -1$ and $x = 3$.The points of intersection are $A(-1, 4)$ and $B(3, 36)$.The required area is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 3$:Area $= \int_{-1}^{3} (8x + 12 - 4x^2) dx$$= [4x^2 + 12x - \frac{4x^3}{3}]_{-1}^{3}$$= (4(9) + 12(3) - \frac{4(27)}{3}) - (4(1) + 12(-1) - \frac{4(-1)}{3})$$= (36 + 36 - 36) - (4 - 12 + \frac{4}{3})$$= 36 - (-8 + \frac{4}{3}) = 36 - (-\frac{20}{3}) = 36 + \frac{20}{3} = \frac{108 + 20}{3} = \frac{128}{3}$ sq. units.

The area (in sq. units) of the region $\{(x, y) \in R^{2} | 4 x^{2} \leq y \leq 8 x+12\}$ is:

Similar Questions

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

The area of the region bounded by the parabola $y=x^2$ and the curve $y=|x|$ is

The area enclosed between the parabolas ${y^2 = 4x}$ and ${x^2 = 4y}$ is

The area of the region $\{(x, y): x^2 \leq y \leq |x^2-4|, y \geq 1\}$ is

The area (in sq. units) bounded by the curves $y=\frac{8}{x}$,$y=2x$ and $x=4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module