CUET Preparation Today
The value of the integral $\int\limits_1^3|x-2| d x$ is: |
2 1 4 5 |
1 |
The correct answer is Option (2) - 1 $\int\limits_1^3|x-2| d x$ $=\int\limits_1^2(2-x)dx+\int\limits_2^3x-2dx$ $=\left|2x-\frac{x^2}{2}\right|_1^2+\left|\frac{x^2}{2}-2x\right|_2^3$ $=4-\frac{4}{2}-2+\frac{1}{2}+\frac{9}{2}-6-\frac{4}{2}+4$ $=1$ |
