$\int\limits_0^2(|x|+|x-2|)dx =$ |
8 4 2 $\frac{1}{2}$ |
4 |
The correct answer is Option (2) → 4 Integral: $\displaystyle \int_{0}^{2} \left(|x| + |x-2|\right)\,dx$ On $[0,2]$: $|x| = x$ (since $x\ge 0$) $|x-2| = 2 - x$ (since $x-2\le 0$) So the integrand becomes: $x + (2 - x) = 2$ Integral: $\displaystyle \int_{0}^{2} 2\,dx = 2[x]_{0}^{2} = 2(2 - 0) = 4$ Answer = 4 |