Question

What is the integral of e^(-4x)dx from 0 to 1?

Answer 1

   
$\texttt{We use the formula: } \\ \int ae^{bx} = abe^{bx} ~~~ a,b \in R\\ \\ a=1 \texttt{ and } b=-4\\ \\ \Longrightarrow ~~ \int\limits^0_1 {e^{-4x}} \, dx = -4e^{-4x}\Big|_0^1 = -4e^{-4\cdot 1} - (-4e^{-4\cdot 0} )= \\ \\ = -4e^{-4} +4e^0 = -4e^{-4} +4e^0 = \boxed{-4e^{-4} +4}$

Answer 2

The integral of e^(-4x)dx from 0 to 1 is determined by first taking the integral of the e^(-4x)dx. The integral of e^(-4x)dx is -1/4 e^-4x. We substitute 0 to this function: This is equal to -1/4. When substituting with 1, the answer is -1/4 e^-4. The answer is 0.25e^4 -1.