Q. 37 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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,