Practicing Success
Practicing Success
CUET Preparation Today
The value of $\int\limits^{1}_{-1}x^2[x]dx$ is : |
$\frac{1}{3}$ $\frac{2}{3}$ 1 $-\frac{1}{3}$ |
$-\frac{1}{3}$ |
The correct answer is Option (4) → $-\frac{1}{3}$ $\int\limits^{1}_{-1}x^2[x]dx$ $=+\int\limits^{0}_{-1}x^2[x]dx+\int\limits^{1}_{0}x^2[x]dx$ $=-\int\limits^{0}_{-1}x^2dx+0$ $=\left[-\frac{x^3}{3}\right]^{0}_{-1}=-\frac{1}{3}$ |
