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.

${\int }_{-1}^3\left|4-x^2\right|dx$

Short answer

The value of integral is $\frac{34}{3}$and the plot is

Step-by-step solution

Step 1. Given information

An expression is given as${\int }_{-1}^3\left|4-x^2\right|dx$

Step 2. Evaluating integral

Simplify the expression as,

$$\begin{gathered} {\int }_{-1}^3\left|4-x^2\right|dx \\ ={\int }_{-1}^{0}4-x^{2}dx+{\int }_0^2{x}^2-4dx+{\int }_2^{3}4-x^{2}dx \\ ={\int }_{-1}^{0}4dx-{\int }_{-1}^0{x}^{2}dx+{\int }_0^2{x}^{2}dx-{\int }_0^{2}4dx-{\int }_2^{3}4dx+{\int }_2^3{x}^{2}dx \\ =4[x{]}_{-1}^0-{\left[\frac{x^3}{3}\right]}_{-1}^0+{\left[\frac{x^3}{3}\right]}_0^2-4[x{]}_0^2-4[x{]}_2^3+{\left[\frac{x^3}{3}\right]}_2^3 \\ =4(2+1)-\left(\frac{8}{3}+\frac{1}{3}\right)-4(3-2)+\left(\frac{27}{3}-\frac{8}{3}\right) \\ =12-\frac{9}{3}-4+\frac{19}{3} \\ =8+\frac{10}{3} \\ =\frac{24+10}{3} \\ =\frac{34}{3} \end{gathered}$$

And the plot is