Integrals — Maths STD 12 Science — Question

Putting x + 2 = 0

$ \Rightarrow x = - 2 \in \left( { - 5,5} \right)$

$\therefore$ From eq. (i),

$I = \int\limits_{ - 5}^{ - 2} {\left| {x + 2} \right|} dx + \int\limits_{ - 2}^5 {\left| {x + 2} \right|dx} $

$= \int\limits_{ - 5}^{ - 2} { - \left( {x + 2} \right)dx + \int\limits_{ - 2}^5 {\left( {x + 2} \right)dx} } $

$= - \left( {\frac{{{x^2}}}{2} + 2x} \right)_{ - 5}^{ - 2} + \left( {\frac{{{x^2}}}{2} + 2x} \right)_{ - 2}^5$

$= - \left[ {\left( {\frac{4}{2} - 4} \right) - \left( {\frac{{25}}{2} - 10} \right)} \right] $ $+ \left[ {\left( {\frac{{25}}{2} + 10} \right) - \left( {\frac{4}{2} - 4} \right)} \right]$

$= - \left( { - 2 - \frac{5}{2}} \right) + \left( {\frac{{45}}{2} + 2} \right)$

$= 2 + \frac{5}{2} + \frac{{45}}{2} + 2$

$ = 4 + 25 = 29$