Answer
2
Work Step by Step
Let \[I=\int_{0}^{4}|\sqrt x-1|dx\]
\[|\sqrt x-1|=\left\{\begin{array}{cc} -(\sqrt x-1)\;\;\;\;
: 0\leq x<1\\
\sqrt x-1\;\;\; :1\leq x\leq4\end{array}\right.\]
\[\Rightarrow I=-\int_{0}^{1}(\sqrt x-1)dx+\int_{1}^{4}(\sqrt x-1)dx\]
\[I=-\left[\frac{2x^{\frac{3}{2}}}{3}-x\right]_{0}^{1}+\left[\frac{2x^{\frac{3}{2}}}{3}-x\right]_{1}^{4}\]
\[I=-\left[\frac{2}{3}-1\right]+\left[\frac{16}{3}-4-\left(\frac{2}{3}-1\right)\right]\]
\[I=-\frac{2(2)}{3}+2(1)-4+\frac{16}{3}\]
\[I=4-4+2=2\]
Hence $I=2$.
