Correct Answer - Option 3 : e - 2
Concept:
Integration by parts
when u and v are functions of x.
\(\smallint u\times v~dx = u\smallint vdx - \smallint \left[ {\frac{{du}}{{dx}}\smallint vdx} \right]dx\)
Integral of standard function
\(\smallint e^{{ax}} dx=\frac{{e^{ax}}}{{a}}\)
Derivative of standard function
\(\frac{{d}}{{dx}}(x^n)=nx^{n-1}\)
Calculation:
\(I = \mathop \smallint \limits_0^1 {x^2}{e^x}dx\)
Using integration by parts where, u = x2 and v= ex
\(I= {x^2}\mathop \smallint {e^x}dx - \smallint \left[ {\frac{d}{{dx}}{x^2}\smallint {e^x}dx} \right]dx\)
\(I = {x^2}{e^x} - \smallint \left[ {2x\times {e^x}} \right]dx\)
Again using integration by parts in \(\smallint {2x\times {e^x}} dx\) where, u = 2x and v = ex
\(I = {x^2}{e^x} - \left( {2x\smallint {e^x}dx - \smallint \left[ {\frac{d}{{\;dx}}2x\smallint {e^x}dx} \right]dx} \right)\)
\(I = {x^2}{e^x} - (2x{e^x} - \smallint 2{e^x}dx)\)
\(I = \left. {\left( {{x^2}{e^x} - 2x{e^x} + 2{e^x}} \right)} \right|_0^1\)
\(I = \left( {{{\left( 1 \right)}^2}{e^{\left( 1 \right)}} - 2\left( 1 \right){e^{\left( 1 \right)}} + 2{e^{\left( 1 \right)}}} \right) - \left( {{{\left( 0 \right)}^2}{e^{\left( 0 \right)}} - 2\left( 0 \right){e^{\left( 0 \right)}} + 2{e^{\left( 0 \right)}}} \right)\)
\(I = \left( {e - 2e + 2e} \right) - \left( {0 - 0 + 2} \right)\)
\(I = e - 2\)
