The area bounded by y = -x^2 + 2x + 3 and y = | vedclass

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

Question

(B) Given the curve $y = -x^2 + 2x + 3$ and the line $y = 0$.To find the points of intersection,set $y = 0$:$-x^2 + 2x + 3 = 0$$x^2 - 2x - 3 = 0$$(x -

Vedclass · Accepted Answer

$\frac{32}{3}$

Vedclass · Answer

(B) Given the curve $y = -x^2 + 2x + 3$ and the line $y = 0$.To find the points of intersection,set $y = 0$:$-x^2 + 2x + 3 = 0$$x^2 - 2x - 3 = 0$$(x - 3)(x + 1) = 0$So,$x = -1$ and $x = 3$.The required area $A$ is given by the integral:$A = \int_{-1}^{3} (-x^2 + 2x + 3) dx$$A = \left[ -\frac{x^3}{3} + x^2 + 3x \right]_{-1}^{3}$$A = \left( -\frac{27}{3} + 9 + 9 \right) - \left( -\frac{(-1)^3}{3} + (-1)^2 + 3(-1) \right)$$A = (-9 + 9 + 9) - \left( \frac{1}{3} + 1 - 3 \right)$$A = 9 - \left( \frac{1 - 6}{3} \right) = 9 - \left( -\frac{5}{3} \right) = 9 + \frac{5}{3} = \frac{27 + 5}{3} = \frac{32}{3}$ square units.

The area bounded by $y = -x^2 + 2x + 3$ and $y = 0$ is

Similar Questions

The area enclosed between the curves $y=x|x|$ and $y=x-|x|$ is :

$AOB$ is the positive quadrant of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ in which $OA=5, OB=3$. The area between the arc $AB$ and the chord $AB$ of the ellipse in sq. units is

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \pi$ is

The area (in square units) of the region bounded by $x^2=8y$,$x=4$ and the $X$-axis is:

Area enclosed by the ellipse $9x^2 + 4y^2 = 1$ in the first quadrant is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module