Question

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

${\int }_2^4 \frac{1}{4-3x}dx$

Short answer

Ans: The exact value of,${\int }_2^4 \frac{1}{4-3x}dx=-\frac{1}{3}(\ln 8)+\frac{1}{3}(\ln 2)$

Step-by-step solution

Step 1. Given information.

given,

${\int }_2^4 \frac{1}{4-3x}dx$

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

The exact value is calculated as shown below,

$\begin{array}{r}{\int }_2^4 \frac{1}{4-3x}dx\\ =-\frac{1}{3}[\ln (|4-3x|){]}_2^4\\ =-\frac{1}{3}[\ln (|4-3(4\left)\right|)-\ln (|4-3(2\left)\right|\left)\right]\\ =-\frac{1}{3}[\ln (8)-\ln (2\left)\right]\\ =-\frac{1}{3}(\ln 8)+\frac{1}{3}(\ln 2)\end{array}$

Therefore, the exact value islocalid="1648629053226" $-\frac{1}{3}(\ln 8)+\frac{1}{3}(\ln 2)$

Step 3. Check 

The required graph is,