The area bounded by the lines y=x,x=-1,x=2 an | vedclass

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

Question

(C) The required area is the sum of the absolute values of the integrals over the intervals where the function is negative and positive.$A = \int_{-1}

Vedclass · Accepted Answer

$\frac{5}{2}$ sq. units

Vedclass · Answer

(C) The required area is the sum of the absolute values of the integrals over the intervals where the function is negative and positive.$A = \int_{-1}^{0} |x| dx + \int_{0}^{2} |x| dx$Since $x  0$ for $x \in [0, 2]$,we have:$A = \int_{-1}^{0} (-x) dx + \int_{0}^{2} (x) dx$$A = \left[ -\frac{x^2}{2} ight]_{-1}^{0} + \left[ \frac{x^2}{2} ight]_{0}^{2}$$A = (0 - (-\frac{(-1)^2}{2})) + (\frac{2^2}{2} - 0)$$A = \frac{1}{2} + 2 = \frac{5}{2} 	ext{ sq. units}$

The area bounded by the lines $y=x$,$x=-1$,$x=2$ and the $X$-axis is

Similar Questions

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

The area of the region bounded by the curve $y = \cos x$,$x = \frac{\pi}{2}$,$x = \frac{3\pi}{2}$,and $y = 0$ is . . . . . . sq. units.

Area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

Let $A$ denote the area bounded by the curve $y=\frac{1}{x}$ and the lines $y=0, x=1, x=10$. Let $B=1+\frac{1}{2}+\ldots+\frac{1}{9}$ and $C=\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{10}$. Then,

Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module