Question

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

${\int }_{-1}^1 \frac{2^x}{4^x}dx$

Short answer

Ans: The exact value of,${\int }_{-1}^1 \frac{2^x}{4^x}dx=\frac{3}{2\ln \left(2\right)}$.

Step-by-step solution

Step 1. Given information.

given,

${\int }_{-1}^1 \frac{2^x}{4^x}dx$

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

The exact value is calculated as shown below,

$\begin{array}{r}{\int }_{-1}^1 \frac{2^x}{4^x}dx\\ ={\int }_{-1}^1 2^x{4}^{-x}dx\\ ={\int }_{-1}^1 2^x{2}^{-2x}dx\\ ={\int }_{-1}^1 2^{-x}dx\\ ={\left[-\frac{1}{\ln \left(2\right)}{2}^{-x}\right]}_{-1}^1\\ ={\left[-\frac{1}{2^x\ln \left(2\right)}\right]}_{-1}^1\\ =\left[-\frac{1}{2^1\ln \left(2\right)}+\frac{1}{2^{-1}\ln \left(2\right)}\right]\\ =\frac{3}{2\ln \left(2\right)}\end{array}$

Therefore, the value is$\frac{3}{2\ln \left(2\right)}$.

Step 3. Check

The required graph is,