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Lelechka [254]
4 years ago
9

How do you integrate xe^(2x)dx?

Mathematics
1 answer:
vesna_86 [32]4 years ago
3 0
You integrate it by parts:


\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx\qquad\quad(i)}\\\\\\ \begin{array}{lcl} \mathsf{u=x}&~\Rightarrow~&\mathsf{du=dx}\\\\ \mathsf{dv=e^{2x}\,dx}&~\Leftarrow~&\mathsf{v=\dfrac{1}{2}\,e^{2x}} \end{array} \end{array}


\large\begin{array}{l} \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=x\cdot \frac{1}{2}\,e^{2x}-\int\!\frac{1}{2}\,e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\int\!e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\int\!\dfrac{1}{2}\cdot 2e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\cdot \frac{1}{2}\int\!e^{2x}\cdot 2\,dx} \end{array}

\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\int\!e^{2x}\cdot 2\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\int\!e^w\,dw\qquad(w=2x)}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\,e^w+C}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\,e^{2x}+C} \end{array}}\qquad\checkmark \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2154227 </span>


\large\textsf{I hope it helps. :-)}


Tags: <em>integrate indefinite integral product by parts polynomial exponential composite substitution integral calculus</em>

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