The area bounded by the curve y = x^2 + 2, th | vedclass

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

Question

(C) The area bounded by the curve $y = f(x)$, the $x$-axis, and the lines $x = a$ and $x = b$ is given by the definite integral $\int_a^b y \, dx$.Her

Vedclass · Accepted Answer

$\frac{13}{3} 	ext{ sq unit}$

Vedclass · Answer

(C) The area bounded by the curve $y = f(x)$, the $x$-axis, and the lines $x = a$ and $x = b$ is given by the definite integral $\int_a^b y \, dx$.Here, $y = x^2 + 2$, $a = 1$, and $b = 2$.$	ext{Required Area} = \int_1^2 (x^2 + 2) \, dx$$= \left[ \frac{x^3}{3} + 2x ight]_1^2$$= \left( \frac{2^3}{3} + 2(2) ight) - \left( \frac{1^3}{3} + 2(1) ight)$$= \left( \frac{8}{3} + 4 ight) - \left( \frac{1}{3} + 2 ight)$$= \left( \frac{8 + 12}{3} ight) - \left( \frac{1 + 6}{3} ight)$$= \frac{20}{3} - \frac{7}{3}$$= \frac{13}{3} 	ext{ sq unit}$.

The area bounded by the curve $y = x^2 + 2$, the $x$-axis, and the lines $x = 1$ and $x = 2$ is:

Similar Questions

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

The area of the region bounded by the curve $y = \cos x$,$-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ and the $X$-axis is . . . . . . sq. units.

The area of the region bounded by the curve $y = \sin 2x$,the $x$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

The area bounded by the curve $y = x |\sin(\frac{x}{2})|$ and the $X$-axis between the lines $x = 0$ and $x = 4\pi$ (in square units) is: (in $\pi$)

Prove that the curves $y^{2}=4x$ and $x^{2}=4y$ divide the area of the square bounded by $x=0, x=4, y=4$ and $y=0$ into three equal parts.

Vedclass Test Series

Exam Paper Generator

Online Exam Module