Question
If \[f\left( x \right) = \left\{ \begin{array}{l}
\left| x \right| + 1,\, - 1 \le x < 0\\
1 + {\left| x \right|^2},\,0 \le x \le 1
\end{array} \right.\] then $$\int_{ - 1}^1 {f\left( x \right)dx} $$ is equal to :
A.
$$ - \frac{1}{6}$$
B.
$$\frac{{17}}{6}$$
C.
$$ - \frac{{17}}{6}$$
D.
none of these
Answer :
$$\frac{{17}}{6}$$
Solution :
$$\eqalign{
& I = \int_{ - 1}^0 {f\left( x \right)dx} + \int_0^1 {f\left( x \right)dx} \cr
& \,\,\,\,\, = \int_{ - 1}^0 {\left( {\left| x \right| + 1} \right)dx} + \int_0^1 {\left( {1 + {{\left| x \right|}^2}} \right)dx} \cr
& \,\,\,\,\, = \int_{ - 1}^0 {\left( { - x + 1} \right)dx} + \int_0^1 {\left( {1 + {x^2}} \right)dx} \cr
& \,\,\,\,\, = \left[ { - \frac{{{x^2}}}{2} + x} \right]_{ - 1}^0 + \left[ {x + \frac{{{x^3}}}{3}} \right]_0^1 \cr
& \,\,\,\,\, = - \left( { - \frac{1}{2} - 1} \right) + \left( {1 + \frac{1}{3}} \right) \cr
& \,\,\,\,\, = \frac{3}{2} + \frac{4}{3} \cr
& \,\,\,\,\, = \frac{{17}}{6} \cr} $$