The area bounded by the curve y = x|x|, X-axi | vedclass

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

Question

(C) The function $y = x|x|$ is defined as $y = x^2$ for $x \ge 0$ and $y = -x^2$ for $x < 0$.Since we are calculating the area, we take the absolute v

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(C) The function $y = x|x|$ is defined as $y = x^2$ for $x \ge 0$ and $y = -x^2$ for $x Since we are calculating the area, we take the absolute value of the function: $\int_{-1}^1 |x|x|| dx$.This can be split into two intervals: $\int_{-1}^0 |-x^2| dx + \int_{0}^1 |x^2| dx$.Since $|-x^2| = x^2$ and $|x^2| = x^2$, the integral becomes $\int_{-1}^0 x^2 dx + \int_{0}^1 x^2 dx$.Evaluating the integrals: $[\frac{x^3}{3}]_{-1}^0 + [\frac{x^3}{3}]_{0}^1$.$= (0 - (-1/3)) + (1/3 - 0) = 1/3 + 1/3 = 2/3$.

The area bounded by the curve $y = x|x|$, $X$-axis and the ordinates $x = -1$ and $x = 1$ is . . . . . . .

Similar Questions

The area of the figure bounded by $y = e^x$,$y = e^{-x}$,and the straight line $x = 1$ is

The area of the region bounded by the lines $y=2x+1$,$y=3x+1$ and $x=4$ 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,

The area (in square units) of the region bounded by the curve $y = |\sin 2x|$ and the $X$-axis in the interval $[0, 2\pi]$ is:

The area bounded by the curve $x^2 = 8y$ and the straight line $x - 8y + 2 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module