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.

${\int }_0^6\left|\right(x-1\left)\right(x-4\left)\right|dx$

Short answer

The value of integral is $15$and the plot is

Step-by-step solution

Step 1. Given information

An expression is given as${\int }_0^6\left|\right(x-1\left)\right(x-4\left)\right|dx$

Step 2. Evaluating integral

We have to change the integral as

$f\left(x\right)=\left\{\begin{array}{lc}(x-1)(x-4)& 0\le x\le 1\\ -(x-1)(x-4)& 1\le x\le 4\\ (x-1)(x-4)& 4\le x\le 6\end{array}\right.$

Now evaluate the integral,

$$\begin{gathered} {\int }_0^6\left|\right(x-1\left)\right(x-4\left)\right|dx={\int }_0^1(x-1)(x-4)dx+{\int }_1^4-(x-1)(x-4)dx+{\int }_4^6(x-1)(x-4)dx \\ ={\int }_0^1\left(x^2-5x+4\right)dx-{\int }_1^4\left(\left(x^2-5x+4\right)\right)dx+{\int }_4^6\left(x^2-5x+4\right)dx \\ ={\left(\frac{x^3}{3}-\frac{5}{2}{x}^2+4x\right)}_0^1-{\left(\frac{x^3}{3}-\frac{5}{2}{x}^2+4x\right)}_1^4+{\left(\frac{x^3}{3}-\frac{5}{2}{x}^2+4x\right)}_4^6 \\ =\left(\frac{1^3}{3}-\frac{5}{2}(1{)}^2+4(1)-0\right)-\left(\left(\frac{4^3}{3}-\frac{5}{2}(4{)}^2+4(4)\right)-\left(\frac{1^3}{3}-\frac{5}{2}(1{)}^2+4(1)\right)\right)+\left(\left(\frac{6^3}{3}-\frac{5}{2}(6{)}^2+4(6)\right)-\left(\frac{4^3}{3}-\frac{5}{2}(4{)}^2+4(4)\right)\right) \\ =\frac{11}{6}-\left(-\frac{9}{2}\right)+\frac{26}{3} \\ =\frac{11}{6}+\frac{9}{2}+\frac{26}{3} \\ =\frac{90}{6} \\ =15 \end{gathered}$$

And the plot is