Question

Evaluate the definite integral \(\int_{{\rm{ - \pi /4}}}^{{\rm{\pi /4}}} {\left( {{{\rm{x}}^{\rm{3}}}{\rm{ + }}{{\rm{x}}^{\rm{4}}}{\rm{tanx}}} \right)} {\rm{dx}}\)

Short answer

Evaluation of symmetry is\({\rm{0}}\).

Step-by-step solution

 Evaluate\({\rm{f( - x)}}\).

So \({\rm{f(x) = }}{{\rm{x}}^{\rm{3}}}{\rm{ + }}{{\rm{x}}^{\rm{4}}}{\rm{tanx,}}\)

\(\begin{array}{c}{\rm{f( - x) = ( - x}}{{\rm{)}}^{\rm{3}}}{\rm{ + ( - x}}{{\rm{)}}^{\rm{4}}}{\rm{tan( - x) }}\\{\rm{ = - }}{{\rm{x}}^{\rm{3}}}{\rm{ + }}{{\rm{x}}^{\rm{4}}}{\rm{tan( - x) }}\\{\rm{ = - }}{{\rm{x}}^{\rm{3}}}{\rm{ - }}{{\rm{x}}^{\rm{4}}}{\rm{tanx}}\;\;\;{\rm{(tan( - x) = - tanx)}}\end{array}\)

 Odd function \(_{{\rm{ - a}}}^{\rm{a}}{\rm{f(x)dx = 0}}\)because symmetry.

So thatsymmetry is \(_{{\rm{ - \pi /4}}}^{{\rm{\pi /4}}}\left( {{{\rm{x}}^{\rm{3}}}{\rm{ + }}{{\rm{x}}^{\rm{4}}}{\rm{tanx}}} \right){\rm{dx = 0}}\)

Hence the define integral\(\int_{{\rm{ - \pi /4}}}^{{\rm{\pi /4}}} {\left( {{{\rm{x}}^{\rm{3}}}{\rm{ + }}{{\rm{x}}^{\rm{4}}}{\rm{tanx}}} \right)} {\rm{dx = 0}}\)