Blog: `int_0^1 (x^2 + 1)e^(-x) dx` Evaluate the integral

Thursday, June 4, 2015

$\displaystyle{\int_{{0}}^{{1}}}{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{x}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{1}}}{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral,using the method of integration by parts

$\displaystyle\int{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}={\left({{x}}^{{2}}+{1}\right)}\int{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}-\int{\left(\frac{{d}}{{\left.{d}{x}}}{\left({{x}}^{{2}}+{1}\right)}\int{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\left({{x}}^{{2}}+{1}\right)}{\left(-{{e}}^{{-{{x}}}}\right)}-\int{\left({2}{x}\cdot{\left(-{{e}}^{{-{{x}}}}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle=-{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}+{2}\int{x}{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}$

applying again integration by parts,

$\displaystyle=-{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}+{2}{\left({x}\cdot\int{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}-\int{\left(\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}\int{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}\right.}$

$\displaystyle=-{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}+{2}{\left({x}{\left(-{{e}}^{{-{{x}}}}\right)}-\int{\left(-{{e}}^{{-{{x}}}}\right)}{\left.{d}{x}\right.}\right)}$

$\displaystyle=-{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}+{2}{\left(-{x}{{e}}^{{-{{x}}}}+\int{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}\right)}$

$\displaystyle=-{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}+{2}{\left(-{x}{{e}}^{{-{{x}}}}+{\left(-{{e}}^{{-{{x}}}}\right)}\right)}$

$\displaystyle=-{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}-{2}{x}{{e}}^{{-{{x}}}}-{2}{{e}}^{{-{{x}}}}$

$\displaystyle=-{{e}}^{{-{{x}}}}{\left({{x}}^{{2}}+{1}+{2}{x}+{2}\right)}$

$\displaystyle=-{{e}}^{{-{{x}}}}{\left({{x}}^{{2}}+{2}{x}+{3}\right)}$

adding a constant to the solution,

$\displaystyle=-{{e}}^{{-{{x}}}}{\left({{x}}^{{2}}+{2}{x}+{3}\right)}+{C}$

Now evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{1}}}{\left({{x}}^{{2}}+{1}\right)}{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}={{\left[-{{e}}^{{-{{x}}}}{\left({{x}}^{{2}}+{2}{x}+{3}\right)}\right]}_{{0}}^{{1}}}$

$\displaystyle={\left[-{{e}}^{{-{{1}}}}{\left({{1}}^{{2}}+{2}\cdot{1}+{3}\right)}\right]}-{\left[-{{e}}^{{0}}{\left({{0}}^{{2}}+{2}\cdot{0}+{3}\right)}\right]}$