Q. 59 - Calculus

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.

${\int }_{-2}^5|x-2|dx$

Step-by-Step Solution

Answer

The value of integral is $\frac{25}{2}$ and the plot is

Step 1. Given information

An expression is given as ${\int }_{-2}^5|x-2|dx$

Step 2. Evaluating integral

Simplify the integral as,

$$\begin{gathered} {\int }_{-2}^5|x-2|dx \\ ={\int }_{-2}^{2}2-xdx+{\int }_2^{5}x-2dx \\ =-{\int }_{-2}^{2}xdx+2{\int }_{-2}^{2}dx+{\int }_2^{5}xdx-2{\int }_2^{5}dx \\ =-{\left[\frac{x^2}{2}\right]}_{-2}^2+2[x{]}_{-2}^2+{\left[\frac{x^2}{2}\right]}_2^5-2[x{]}_2^5 \\ =-\frac{4}{2}+\frac{4}{2}+2(2+2)+\frac{25}{2}-\frac{4}{2}-2[5-2] \\ =8+\frac{21}{2}-6 \\ =8+\frac{21-12}{2} \\ =8+\frac{9}{2} \\ =\frac{25}{2} \end{gathered}$$

And the plot for verification is