Q. 40 - 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. 

   ${\int }_{-3}^2 \frac{1}{(x+5{)}^2}dx$

Step-by-Step Solution

Answer

Ans:    The exact value is, ${\int }_{-3}^2 \frac{1}{(x+5{)}^2}dx =\frac{5}{14}$

Step 1. Given information.

given expression,

          ${\int }_{-3}^2 \frac{1}{(x+5{)}^2}dx$

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

The exact value is calculated as shown below,  

      $\begin{array}{r}{\int }_{-3}^2 \frac{1}{(x+5{)}^2}dx\\ ={\int }_{-3}^2 (x+5{)}^{-2}dx\\ ={\left[\frac{(x+5{)}^{-2+1}}{-2+1}\right]}_{-3}^2\\ ={\left[\frac{(x+5{)}^{-1}}{-1}\right]}_{-3}^2\\ ={\left[\frac{-1}{(x+5)}\right]}_{-3}^2\\ =\frac{-1}{(2+5)}+\frac{1}{(-3+5)}\\ =\frac{-1}{7}+\frac{1}{2}\\ =\frac{-2+7}{14}\\ =\frac{5}{14}\end{array}$

Therefore, the exact value is $\frac{5}{14}$.$\frac{5}{14}$

Step 3. Check:

The required graph is,