evaluate the integral below. ∫₁⁴ 2xe³ˣ dx be sure to place the argument of any trigonometric or logarithmic…

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$. Let $u = 2x$ and $dv=e^{3x}dx$. Then $du = 2dx$ and $v=\frac{1}{3}e^{3x}$.

Step2: Apply the integration - by - parts formula

$\int_{1}^{4}2xe^{3x}dx=\left[2x\cdot\frac{1}{3}e^{3x}\right]{1}^{4}-\int{1}^{4}\frac{1}{3}e^{3x}\cdot2dx$. First, evaluate $\left[2x\cdot\frac{1}{3}e^{3x}\right]{1}^{4}=\frac{2}{3}\times4e^{12}-\frac{2}{3}\times1e^{3}=\frac{8e^{12}}{3}-\frac{2e^{3}}{3}$. Second, evaluate $\int{1}^{4}\frac{2}{3}e^{3x}dx=\frac{2}{3}\times\frac{1}{3}\left[e^{3x}\right]_{1}^{4}=\frac{2}{9}(e^{12}-e^{3})$.

Step3: Simplify the result

$\int_{1}^{4}2xe^{3x}dx=\frac{8e^{12}}{3}-\frac{2e^{3}}{3}-\frac{2e^{12}}{9}+\frac{2e^{3}}{9}$. Combining like - terms: [ \begin{align*} &=\left(\frac{8e^{12}}{3}-\frac{2e^{12}}{9}\right)+\left(-\frac{2e^{3}}{3}+\frac{2e^{3}}{9}\right)\ &=\frac{24e^{12}-2e^{12}}{9}+\frac{- 6e^{3}+2e^{3}}{9}\ &=\frac{22e^{12}}{9}-\frac{4e^{3}}{9} \end{align*} ]

Answer:

$\frac{22e^{12}-4e^{3}}{9}$