Question

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer.

${\int }_0^1 \frac{1}{2e^x}dx$

Short answer

Ans: The exact value of${\int }_0^1 \frac{1}{2e^x}dx=-\frac{1}{2e}+\frac{1}{2}$

Step-by-step solution

Step 1. Given information.

given,

${\int }_0^1 \frac{1}{2e^x}dx$

Step 2. The objective is to determine the exact value of the definite integral.   

The exact value is calculated as shown below,

$\begin{array}{r}{\int }_0^1 \frac{1}{2e^x}dx\\ =\frac{1}{2}{\int }_0^1 \frac{1}{e^x}dx\\ =\frac{1}{2}{\int }_0^1 e^{-x}dx\\ =\frac{1}{2}{\left[-e^{-x}\right]}_0^1\\ =-\frac{1}{2}\left[e^{-1}-e^0\right]\\ =-\frac{1}{2e}+\frac{1}{2}\end{array}$

Therefore, the exact value is, $-\frac{1}{2e}+\frac{1}{2}$

Step 3. Check

The required graph is,