Q 19. - Calculus

Question

Each expression that follows is the result of a calculation that uses integration by parts. That is, each is an expression of the form uv −  v du for some functions u and v. Identify u, v, du, and dv, and determine the original integral  u dv.

$\frac{x{2}^x}{\ln 2}-\frac{1}{\ln 2}\int 2^x dx$

Step-by-Step Solution

Answer

$\begin{array}{rcl}u& =& x\\ v& =& \frac{2^x}{\ln 2}\\ du& =& dx\\ dv& =& 2^x dx\\ \int u dv& =& \int x\left(2^x\right)dx\end{array}$

Step 1. Given information

Integral is $\frac{x{2}^x}{\ln 2}-\frac{1}{\ln 2}\int 2^x dx$

Step 2. Explanation

Take $u=x,v=\frac{2^x}{\ln 2}$

$\begin{array}{rcl}du& =& dx\\ dv& =& \frac{\ln 2}{\ln 2}{2}^x dx\\ & =& 2^x\\ \int u dv& =& \int x{2}^x dx\end{array}$