Question

Solve the integral:$\int \left(3-x\right)e^{2x}dx$

Short answer

The required answer is$\frac{e^{2x}}{2}\left(-x+3\right)+\frac{1}{4}{e}^{2x}+c$.

Step-by-step solution

Step 1. Given information. 

We have given integral is$\int \left(3-x\right)e^{2x}dx$.

Step 2. Solve the integration by parts .  

We have, $u=-x+3,dv=e^{2x}dx$

$$\begin{gathered} u=-x+3 \\ du=-dx \end{gathered}$$

and

$$\begin{gathered} dv=e^{2x}dx \\ v=\int e^{2x}dx \\ v=\frac{e^{2x}}{2} \end{gathered}$$

The formula of integration by parts is $\int udv=uv-\int vdu$

$$\begin{gathered} \int \left(3-x\right)e^{2x}dx \\ =\left[\left(-x+3\right)\left(\frac{e^{2x}}{2}\right)-\int -\left(\frac{e^{2x}}{2}\right)dx\right] \\ =(-x+3)\left(\frac{e^{2x}}{2}\right)+\frac{1}{2}\int e^{2x}dx \end{gathered}$$

Step 3. Integration by parts. 

We have,

$$\begin{gathered} u=2x \\ du=2dx \\ \frac{du}{2}=dx \end{gathered}$$

So,

$$\begin{gathered} \frac{e^{2x}}{2}(-x+3)+\frac{1}{2}\int \frac{e^{2x}}{2}du \\ =\frac{e^{2x}}{2}(-x+3)+\frac{1}{4}\int e^{2x}du \\ =\frac{e^{2x}}{2}(-x+3)+\frac{1}{4}{e}^{2x}+c \end{gathered}$$