Chapter 6.2, Problem 38E

(a)

To determine

To Find: Points of discontinuity of the integrand on the interval of integration.

Answer to Problem 38E

Points of discontinuity of the integrand on the interval of integration is $x=–6,–5,–4,–3,–2,–1,0,1,2,3,4,5$

Explanation of Solution

Given information:

  ${{\displaystyle \int }}_{-6}^{5}2\mathrm{int}(x-3)dx$

Calculation:

Any greatest integer function is discontinuous at all integers.

So,

  $f\left(x\right)=2int\left(x–3\right)$ , it is discontinuous at

  $x=–6,–5,–4,–3,–2,–1,0,1,2,3,4,5$

(b)

To determine

To Evaluate: the given integral using the area.

Answer to Problem 38E

The integral using the area is $–88$

Explanation of Solution

Given information:

  ${{\displaystyle \int }}_{-6}^{5}2\mathrm{int}(x-3)dx$

Calculation:

For any integral function,

  ${{\displaystyle \int }}_a^{b}f(x)dx=$ (Area above the x-axis)− (area below the x-axis)

  

So,

  $\begin{array}{l}{{\displaystyle \int }}_{-6}^{5}2\mathrm{int}(x-3)dx=(sumofareaofrectanglesabovethex-axis)–\hfill \\ \left(sumofareaofrectanglesbelowthex-axis\right)\hfill \\ =\left(0\times 1+2\times 1\right)–\hfill \\ \left(18\times 1+14\times 1+12\times 1+10\times 1+8\times 1+6\times 1+4\times 1+2\times 1\right)\hfill \\ =2–90\hfill \\ =–88\hfill \end{array}$