Find the area of the region bounded by the cu | vedclass

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

Question

(A) The area bounded by the curves $y=x^{2}+2, \,y=x, \,x=0,$ and $x=3$ is given by the integral of the difference between the upper curve and the low

Vedclass · Accepted Answer

$\frac{21}{2}$

Vedclass · Answer

(A) The area bounded by the curves $y=x^{2}+2, \,y=x, \,x=0,$ and $x=3$ is given by the integral of the difference between the upper curve and the lower curve within the given limits.Area $= \int_{0}^{3} ((x^{2}+2) - x) \, dx$$= \int_{0}^{3} (x^{2} - x + 2) \, dx$$= \left[ \frac{x^{3}}{3} - \frac{x^{2}}{2} + 2x ight]_{0}^{3}$$= \left( \frac{3^{3}}{3} - \frac{3^{2}}{2} + 2(3) ight) - (0)$$= \left( \frac{27}{3} - \frac{9}{2} + 6 ight)$$= 9 - 4.5 + 6$$= 10.5 = \frac{21}{2} 	ext{ sq. units}$

Find the area of the region bounded by the curves $y=x^{2}+2, \,y=x,\, x=0$ and $x=3$.

Similar Questions

The area bounded by the parabola $y^2=4ax$ and its latus-rectum $x=a$ is

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$.

The area of the region bounded by the $x$-axis and the curves defined by $y = \tan x$ for $(-\pi/3 \le x \le \pi/3)$ is

The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and its latus rectum is . . . . . . sq. units. (where $e$ denotes the eccentricity of the ellipse).

The area of the region enclosed between the parabola $y^{2}=x$ and the line $y=mx$ is $\frac{1}{48}$. Then, the value of $m$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module