Answer
$$A = \frac{{{2^{5/2}} - {2^{3/2}} - 1}}{3}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {x + 1} - 1;\,\,\,\left[ { - 1,1} \right] \cr
& {\text{From the graph we can see that the area is given by}} \cr
& A = \int_{ - 1}^0 {\left( {1 - \sqrt {x + 1} } \right)} dx + \int_0^1 {\left( {\sqrt {x + 1} - 1} \right)dx} \cr
& A = \left[ {x - \frac{{2{{\left( {x + 1} \right)}^{3/2}}}}{3}} \right]_{ - 1}^0 + \left[ {\frac{{2{{\left( {x + 1} \right)}^{3/2}}}}{3} - x} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& A = \left[ {\left( { - 1 - \frac{{2{{\left( { - 1 + 1} \right)}^{3/2}}}}{3}} \right) - \left( {0 - \frac{{2{{\left( {0 + 1} \right)}^{3/2}}}}{3}} \right)} \right] \cr
& \,\,\,\,\,\, + \left[ {\left( {\frac{{2{{\left( {1 + 1} \right)}^{3/2}}}}{3} - 1} \right)} \right] - \left[ {\left( {\frac{{2{{\left( {0 + 1} \right)}^{3/2}}}}{3} - 1} \right)} \right] \cr
& A = \left[ { - 1 + \frac{2}{3}} \right]\, + \left[ {\frac{{{2^{5/2}}}}{3} - 1} \right] - \left[ {\frac{{{2^{3/2}}}}{3} - 1} \right] \cr
& A = - \frac{1}{3} + \frac{{{2^{5/2}}}}{3} - 1 - \frac{{{2^{3/2}}}}{3} + 1 \cr
& A = - \frac{1}{3} + \frac{{{2^{5/2}} - {2^{3/2}}}}{3} \cr
& A = \frac{{{2^{5/2}} - {2^{3/2}} - 1}}{3} \cr} $$
