Question Number 135361 by leena12345 last updated on 12/Mar/21
![f(x)=3x^2 +6x,[−1,5] find−the−average−value](https://www.tinkutara.com/question/Q135361.png)
$${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x},\left[−\mathrm{1},\mathrm{5}\right] \\ $$$${find}−{the}−{average}−{value} \\ $$$$ \\ $$
Answered by mr W last updated on 12/Mar/21
![average of f(x) over [a, b]: f_m =((∫_a ^b f(x)dx)/(b−a))=((∫_(−1) ^5 (3x^2 +6x)dx)/(5−(−1))) =(([x^3 +3x^2 ]_(−1) ^5 )/6)=((5^3 +3×5^2 −(−1)^3 −3(−1)^2 )/6) =((198)/6)=33](https://www.tinkutara.com/question/Q135392.png)
$${average}\:{of}\:{f}\left({x}\right)\:{over}\:\left[{a},\:{b}\right]: \\ $$$${f}_{{m}} =\frac{\int_{{a}} ^{{b}} {f}\left({x}\right){dx}}{{b}−{a}}=\frac{\int_{−\mathrm{1}} ^{\mathrm{5}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}\right){dx}}{\mathrm{5}−\left(−\mathrm{1}\right)} \\ $$$$=\frac{\left[{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} \right]_{−\mathrm{1}} ^{\mathrm{5}} }{\mathrm{6}}=\frac{\mathrm{5}^{\mathrm{3}} +\mathrm{3}×\mathrm{5}^{\mathrm{2}} −\left(−\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}\left(−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{6}} \\ $$$$=\frac{\mathrm{198}}{\mathrm{6}}=\mathrm{33} \\ $$