Q. 20 - 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. (Hint: The integrands that involve absolute values will have to be considered piecewise.) 

${\int }_{-3}^3\left(x-1\right)\left(x+3\right)dx$

Step-by-Step Solution

Answer

The required value is $0$.

Step 1. Given Information

We are given the definite integral ${\int }_{-3}^3\left(x-1\right)\left(x+3\right)dx$ and we need to use the Fundamental Theorem of Calculus to find the exact value of the integral.      

Step 2. Finding the integral

The required value is:   

$$\begin{gathered} {\int }_{-3}^3\left(x-1\right)\left(x+3\right)dx \\ ={\int }_{-3}^3\left(x^2-x+3x-3\right)dx \\ ={\int }_{-3}^3\left(x^2-2x-3\right)dx \\ ={\int }_{-3}^3{x}^{2}dx-{\int }_{-3}^{3}2xdx-{\int }_{-3}^{3}3dx \\ ={\left[\frac{x^3}{3}\right]}_{-3}^3-2{\left[\frac{x^2}{2}\right]}_{-3}^3-{\left[3x\right]}_{-3}^3 \\ =\left[\frac{3^3}{3}-\frac{{\left(-3\right)}^3}{3}\right]-2\left[\frac{3^2}{2}-\frac{{\left(-3\right)}^2}{2}\right]-\left[3\left(3-\left(-3\right)\right)\right] \\ =9+9-9+9-9-9 \\ =0 \end{gathered}$$

Step 3. Rechecking solution

The required graph is: 

After plotting the graph we can see that the area under the graph is exactly same as the area obtained from the definite integral. The area under the graph is $0$ square units.