Chapter 5.6, Problem 1E

To determine

To Calculate: ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}$

Answer to Problem 1E

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}=-18$

Explanation of Solution

Given information:

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}$

Concept and Formula Used:

  ${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C,}{\displaystyle \underset{a}{\overset{b}{\int }}cdx=c\left(b-a\right)}$ ,

Calculation:

It is given that

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}$

  $\text{Using }{\displaystyle \underset{a}{\overset{b}{\int }}\left(f(x)+g(x)\right)dx}={\displaystyle \underset{a}{\overset{b}{\int }}\left(f(x)\right)dx}+{\displaystyle \underset{a}{\overset{b}{\int }}\left(g(x)\right)dx}$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}={\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2\right)dx}-{\displaystyle \underset{1}{\overset{4}{\int }}\left(4x\right)dx}-{\displaystyle \underset{1}{\overset{4}{\int }}\left(3\right)dx}$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}={\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2\right)dx}-4{\displaystyle \underset{1}{\overset{4}{\int }}\left(x\right)dx}-3{\displaystyle \underset{1}{\overset{4}{\int }}dx}$

  $\text{Using }{\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}={\left[\frac{x^{2+1}}{2+1}\right]}_1^4-4{\left[\frac{x^{1+1}}{1+1}\right]}_1^4-3{\left[x\right]}_1^4$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}={\left[\frac{x^3}{3}\right]}_1^4-4{\left[\frac{x^2}{2}\right]}_1^4-3{\left[x\right]}_1^4$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}=\frac{1}{3}\left[4^3-1\right]-2\left[4^2-1\right]-3\left[4-1\right]$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}=\frac{1}{3}\left[63\right]-2\left[15\right]-3\left[3\right]$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}=21-30-9$

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}=-18$

Conclusion:

  ${\displaystyle \underset{1}{\overset{4}{\int }}\left(x^2-4x-3\right)dx}=-18$