Correct Answer - Option 4 : 4
Concept:
- \(\mathop \smallint \nolimits_a^b f\left( x \right)dx = \;\mathop \smallint \nolimits_a^c f\left( x \right)dx + \;\mathop \smallint \nolimits_c^b f\left( x \right)dx\)
- \(\mathop \smallint \nolimits_a^b f'\left( x \right)dx = \;\left[ {f\left( x \right)} \right]_a^b = \left[ {f\left( b \right) - f\left( a \right)} \right]\)
- \(\smallint {x^n}dx = \;\frac{{{x^{n\; + \;1}}}}{{n + 1}} + c\)
Calculation:
Given: \(\mathop \smallint \nolimits_0^3 \left| {x - 2} \right|dx\)
Now, f(x) = \(\left| {x - 2} \right|\; \Rightarrow \;f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - \;\left( {x - 2} \right),\;\;\;0 \le x < 2}\\ {x - 2,\;\;\;2 \le x \le 4} \end{array}} \right.\)
\(\mathop \smallint \nolimits_0^4 \left| {x - 2} \right|dx = \;\mathop \smallint \nolimits_0^2 - \left( {{\rm{x}} - 2} \right)\;dx + \;\mathop \smallint \nolimits_2^4 \left( {{\rm{x}} - 2} \right)\;dx\)
\(\rm \;\mathop \smallint \nolimits_0^2 \left( {2 - x} \right)\;dx + \;\mathop \smallint \nolimits_2^4 \left( {{\rm{x}} - 2} \right)\;dx\)
\( = \;\left[ {2x - \;\frac{{{x^2}}}{2}} \right]_0^2\; + \;\left[ {\frac{{{x^2}}}{2} - 2x} \right]_2^4\;\)
= \(\rm [(4 \ - {\frac {4}{2}})\ -\ 0] \ -\ [({\frac {16}{2}}\ -\ 8)\ -\ ({\frac {4}{2}}\ -\ 4)] \)
= 2 - (- 2)
= 4
