Q. 34 - 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.

    ${\int }_2^4 \frac{1}{4-3x}dx$

Step-by-Step Solution

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 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 is $-\frac{1}{3}(\ln 8)+\frac{1}{3}(\ln 2)$

Step 3. Check

The required graph is,