Question Number 210032 by klipto last updated on 29/Jul/24
$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5}\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{dx}} \\ $$
Answered by Sutrisno last updated on 29/Jul/24
![=∫_6 ^0 2e^((1/3)x) dx+5∫_6 ^0 xe^((1/3)x) dx =6e^((1/3)x) ∣_6 ^0 +5[3xe^((1/3)x) −9e^((1/3)x) ]∣_6 ^0 =6(e^0 −e^2 )+5[(0−9e^0 )−(16e^2 −9e^2 )] =6(1−e^2 )+5[−9−7e^2 ] =6−6e^2 −45−35e^2 =−39−41e^2](https://www.tinkutara.com/question/Q210033.png)
$$=\int_{\mathrm{6}} ^{\mathrm{0}} \mathrm{2}{e}^{\frac{\mathrm{1}}{\mathrm{3}}{x}} {dx}+\mathrm{5}\int_{\mathrm{6}} ^{\mathrm{0}} {xe}^{\frac{\mathrm{1}}{\mathrm{3}}{x}} {dx} \\ $$$$=\mathrm{6}{e}^{\frac{\mathrm{1}}{\mathrm{3}}{x}} \mid_{\mathrm{6}} ^{\mathrm{0}} +\mathrm{5}\left[\mathrm{3}{xe}^{\frac{\mathrm{1}}{\mathrm{3}}{x}} −\mathrm{9}{e}^{\frac{\mathrm{1}}{\mathrm{3}}{x}} \right]\mid_{\mathrm{6}} ^{\mathrm{0}} \\ $$$$=\mathrm{6}\left({e}^{\mathrm{0}} −{e}^{\mathrm{2}} \right)+\mathrm{5}\left[\left(\mathrm{0}−\mathrm{9}{e}^{\mathrm{0}} \right)−\left(\mathrm{16}{e}^{\mathrm{2}} −\mathrm{9}{e}^{\mathrm{2}} \right)\right] \\ $$$$=\mathrm{6}\left(\mathrm{1}−{e}^{\mathrm{2}} \right)+\mathrm{5}\left[−\mathrm{9}−\mathrm{7}{e}^{\mathrm{2}} \right] \\ $$$$=\mathrm{6}−\mathrm{6}{e}^{\mathrm{2}} −\mathrm{45}−\mathrm{35}{e}^{\mathrm{2}} \\ $$$$=−\mathrm{39}−\mathrm{41}{e}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by klipto last updated on 29/Jul/24
$$\mathrm{i}\:\mathrm{guess}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{typo} \\ $$$$ \\ $$
Answered by Frix last updated on 29/Jul/24
![∫_α ^β (ax+b)e^(cx) dx=[((a/c)x−(a/c^2 )+(b/c))e^(cx) ]_α ^β ⇒ Answer is −3(17e^2 +13)](https://www.tinkutara.com/question/Q210040.png)
$$\underset{\alpha} {\overset{\beta} {\int}}\left({ax}+{b}\right)\mathrm{e}^{{cx}} {dx}=\left[\left(\frac{{a}}{{c}}{x}−\frac{{a}}{{c}^{\mathrm{2}} }+\frac{{b}}{{c}}\right)\mathrm{e}^{{cx}} \right]_{\alpha} ^{\beta} \\ $$$$\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:−\mathrm{3}\left(\mathrm{17e}^{\mathrm{2}} +\mathrm{13}\right) \\ $$
Commented by klipto last updated on 29/Jul/24
$$\mathrm{yeah}\:\mathrm{my}\:\mathrm{bro}\:\mathrm{thanks} \\ $$
Answered by klipto last updated on 29/Jul/24
![∫_6 ^0 (2+5x)e^((1/3)x) dx aliter determinant ((D,I),((2+5x),e^((1/3)x) ),((−5),(3e^((1/3)x) )),(0,( 9e^((1/3)x) )))_ ∫_6 ^0 (2+5x)e^((1/3)x) dx=[(2+5x)(3e^((1/3)x) )]_6 ^0 −45(e^((1/3)x) )_6 ^0 ∫_6 ^0 (2+5x)e^((1/3)x) dx=[3((2+5x)e^((1/3)x) )]_6 ^0 −45[e^((1/3)x) ]_6 ^0 ∫_6 ^0 (2+5x)e^((1/3)x) dx=3[(2−32e^2 )]−45[1−e^2 ] ∫_6 ^0 (2+5x)e^((1/3)x) dx=6−96e^2 −45+45e^2 ∫_6 ^0 (2+5x)e^((1/3)x) dx=−39−51e^2 ∫_6 ^0 (2+5x)e^((1/3)x) dx using IBP ∫udv=uv−∫vdu u=2+5x dv=e^((1/3)x) dx{v=3e^((1/3)x) } du=5, m=(1/3)x,(dm/dx)=(1/3),dx=3 ∫_6 ^0 (2+5x)e^((1/3)x) dx=[(2+5x)(3e^((1/3)x) )]_6 ^0 −∫_6 ^0 5(3e^((1/3)x) ) ∫_6 ^0 (2+5x)e^((1/3)x) dx=[3(2+5x)(e^((1/3)x) )]_6 ^0 −15∫_6 ^0 e^((1/3)x) ∫_6 ^0 (2+5x)e^((1/3)x) dx=[3(2+5x)(e^((1/3)x) )]_6 ^0 −15[3e^((1/3)x) ] ∫_6 ^0 (2+5x)e^((1/3)x) dx=3[(2+5x)(e^((1/3)x) )]_6 ^0 −45[e^((1/3)x) ]_6 ^0 ∫_6 ^0 (2+5x)e^((1/3)x) dx=3[(2−32e^2 )]−45[1−e^2 ] ∫_6 ^0 (2+5x)e^((1/3)x) dx=6−96e^2 −45+45e^2 ∫_6 ^0 (2+5x)e^((1/3)x) dx=−39−51e^2 ✓ klipto−quanta.⊎](https://www.tinkutara.com/question/Q210043.png)
$$ \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx} \\ $$$$\boldsymbol{\mathrm{aliter}} \\ $$$$\begin{array}{|c|c|c|c|}{\boldsymbol{\mathrm{D}}}&\hline{\boldsymbol{\mathrm{I}}}\\{\mathrm{2}+\mathrm{5}\boldsymbol{\mathrm{x}}}&\hline{\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} }\\{−\mathrm{5}}&\hline{\mathrm{3}\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} }\\{\mathrm{0}}&\hline{\:\:\mathrm{9}\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} }\\\hline\end{array}_{} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\left[\left(\mathrm{2}+\mathrm{5x}\right)\left(\mathrm{3e}^{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}} \right)\right]_{\mathrm{6}} ^{\mathrm{0}} −\mathrm{45}\left(\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \right)_{\mathrm{6}} ^{\mathrm{0}} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\left[\mathrm{3}\left(\left(\mathrm{2}+\mathrm{5x}\right)\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}} \right)\right]_{\mathrm{6}} ^{\mathrm{0}} −\mathrm{45}\left[\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \right]_{\mathrm{6}} ^{\mathrm{0}} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\mathrm{3}\left[\left(\mathrm{2}−\mathrm{32}\boldsymbol{\mathrm{e}}^{\mathrm{2}} \right)\right]−\mathrm{45}\left[\mathrm{1}−\boldsymbol{\mathrm{e}}^{\mathrm{2}} \right] \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\mathrm{6}−\mathrm{96}\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\mathrm{45}+\mathrm{45}\boldsymbol{\mathrm{e}}^{\mathrm{2}} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=−\mathrm{39}−\mathrm{51}\boldsymbol{\mathrm{e}}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{IBP}} \\ $$$$\int\boldsymbol{\mathrm{udv}}=\boldsymbol{\mathrm{uv}}−\int\boldsymbol{\mathrm{vdu}} \\ $$$$\boldsymbol{\mathrm{u}}=\mathrm{2}+\mathrm{5}\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{dv}}=\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{dx}}\left\{\boldsymbol{\mathrm{v}}=\mathrm{3}\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \right\} \\ $$$$\boldsymbol{\mathrm{du}}=\mathrm{5},\:\:\boldsymbol{\mathrm{m}}=\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}},\frac{\boldsymbol{\mathrm{dm}}}{\boldsymbol{\mathrm{dx}}}=\frac{\mathrm{1}}{\mathrm{3}},\boldsymbol{\mathrm{dx}}=\mathrm{3} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\left[\left(\mathrm{2}+\mathrm{5x}\right)\left(\mathrm{3e}^{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}} \right)\right]_{\mathrm{6}} ^{\mathrm{0}} −\int_{\mathrm{6}} ^{\mathrm{0}} \mathrm{5}\left(\mathrm{3e}^{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}} \right) \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\left[\mathrm{3}\left(\mathrm{2}+\mathrm{5x}\right)\left(\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}} \right)\right]_{\mathrm{6}} ^{\mathrm{0}} −\mathrm{15}\int_{\mathrm{6}} ^{\mathrm{0}} \boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\left[\mathrm{3}\left(\mathrm{2}+\mathrm{5x}\right)\left(\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}} \right)\right]_{\mathrm{6}} ^{\mathrm{0}} −\mathrm{15}\left[\mathrm{3}\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \right] \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\mathrm{3}\left[\left(\mathrm{2}+\mathrm{5x}\right)\left(\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}} \right)\right]_{\mathrm{6}} ^{\mathrm{0}} −\mathrm{45}\left[\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \right]_{\mathrm{6}} ^{\mathrm{0}} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\mathrm{3}\left[\left(\mathrm{2}−\mathrm{32}\boldsymbol{\mathrm{e}}^{\mathrm{2}} \right)\right]−\mathrm{45}\left[\mathrm{1}−\boldsymbol{\mathrm{e}}^{\mathrm{2}} \right] \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=\mathrm{6}−\mathrm{96}\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\mathrm{45}+\mathrm{45}\boldsymbol{\mathrm{e}}^{\mathrm{2}} \\ $$$$\int_{\mathrm{6}} ^{\mathrm{0}} \left(\mathrm{2}+\mathrm{5x}\right)\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{x}}} \mathrm{dx}=−\mathrm{39}−\mathrm{51}\boldsymbol{\mathrm{e}}^{\mathrm{2}} \checkmark \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}}.\biguplus \\ $$