Q 18. - Calculus

Question

For each pair of functions u(x) and v(x) in Exercise, fill in the blanks to complete each of the following:

$u\left(x\right)=x^3,v\left(x\right)=e^{3x}$

(a) $\frac{d}{dx}\left(u\left(x\right)v\left(x\right)\right)=\_\_\_\_\_\_$

(b) width="178" style="max-width: none; vertical-align: -10px;" $\int \_\_\_\_ dx=u\left(x\right)v\left(x\right)+C$

(c) $\int u dv=\_\_\_\_\_$

Step-by-Step Solution

Answer

(a) $3x^3{e}^{3x}+2x^2{e}^{3x}$

(b) $3x^3{e}^{3x}+2x^2{e}^{3x}$

(c) $\left(\frac{x^3}{3}-\frac{x^2}{3}+\frac{2x}{9}-\frac{2}{27}\right)e^{3x}+C$

Step 1. Given information

$u\left(x\right)=x^3,v\left(x\right)=e^{3x}$

Part (a) Step 1. Explanation

$\begin{array}{rcl}\frac{d}{dx}\left(x^3{e}^{3x}\right)& =& x^3\frac{d}{dx}\left(e^{3x}\right)+\left(e^{3x}\right)\frac{d}{dx}\left(x^3\right)\\ & =& \underline{3x^3{e}^{3x}+2x^2{e}^{3x}}\end{array}$

Part (b) Step 1. Explanation

$\int \_\_\_\_ dx=u\left(x\right)v\left(x\right)+C$

As the integration of the value is the anti derivative of the $u\left(x\right)v\left(x\right)$

$\int \underline{\left(3x^3{e}^{3x}+2x^2{e}^{3x}\right)} dx=x^3{e}^{3x}+C$

Part (c) Step 1. Explanation

$\begin{array}{rcl}\int u dv& =& 3\int x^3{e}^{3x} dx\\ & =& 3\left[x^3\frac{e^{3x}}{3}-\int \frac{e^{3x}}{3}3x^2\right]+C\\ & =& 3\left[x^3\frac{e^{3x}}{3}-\int e^{3x}{x}^2\right]+C\\ & =& 3\left[x^3\frac{e^{3x}}{3}-\left(x^2\frac{e^{3x}}{3}-\int \frac{e^{3x}}{3}2x\right)\right]+C\\ & =& \underline{\left(\frac{x^3}{3}-\frac{x^2}{3}+\frac{2x}{9}-\frac{2}{27}\right)e^{3x}+C}\end{array}$