CUET Preparation Today
The value of $\int\limits_0^3|2 x-6| dx$ is: |
$-\frac{9}{4}$ $\frac{19}{2}$ 9 -3 |
9 |
The correct answer is Option (3) → 9 $\int\limits_0^3|x-3|dx$ since from 0 → 3, x ⇒ 3 $⇒I=2\int\limits_0^33-xdx$ $=2\left[3x-\frac{x^2}{2}\right]_0^3$ ⇒ 9 units |
