Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises $19–64$. Use

a graph to check your answer.

${\int }_0^1\frac{2^x-3^x}{4^x}dx$

Short answer

The value of integral is $\left(\frac{1}{2}\left(\frac{1}{\ln \left(\frac{1}{2}\right)}\right)-\frac{1}{\ln \left(\frac{1}{2}\right)}\right)-\left(\frac{3}{4}\left(\frac{1}{\ln \left(\frac{3}{4}\right)}\right)-\frac{1}{\ln \left(\frac{3}{4}\right)}\right)$

And the graph is

Step-by-step solution

Step 1. Given information

An expression is given as${\int }_0^1\frac{2^x-3^x}{4^x}dx$

Step 2. Evaluating integral

Simplify the integral as,

$$\begin{gathered} {\int }_0^1\frac{2^x-3^x}{4^x}dx \\ ={\int }_0^1{\left(\frac{2}{4}\right)}^{x}dx-{\int }_0^1{\left(\frac{3}{4}\right)}^{x}dx \\ ={\int }_0^1{\left(\frac{1}{2}\right)}^{x}dx-{\int }_0^1{\left(\frac{3}{4}\right)}^{x}dx \\ ={\left[\frac{1}{\ln \left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}^x\right]}_0^1-{\left[\frac{1}{\ln \left(\frac{3}{4}\right)}{\left(\frac{3}{4}\right)}^x\right]}_0^1 \\ =\left(\frac{1}{2}\left(\frac{1}{\ln \left(\frac{1}{2}\right)}\right)-\frac{1}{\ln \left(\frac{1}{2}\right)}\right)-\left(\frac{3}{4}\left(\frac{1}{\ln \left(\frac{3}{4}\right)}\right)-\frac{1}{\ln \left(\frac{3}{4}\right)}\right) \end{gathered}$$

And the graph is