Chapter 6.2, Problem 2E

To determine

To express the given limit as a definite integral.

Answer to Problem 2E

  ${\displaystyle \underset{-7}{\overset{5}{\int }}\left(x^2-3x\right)dx}$

Explanation of Solution

Given:

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(c_k^2-3c_k\right)\Delta x},\left[-7,5\right]$

Concept used:

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$

Calculation:

  $\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(c_k^2-3c_k\right)\Delta x},\left[-7,5\right]\\ \Rightarrow f\left(c_k\right)=c_k^2-3c_k\\ \Rightarrow f\left(x\right)=x^2-3x\end{array}$

Therefore definite integral is,

  ${\displaystyle \underset{-7}{\overset{5}{\int }}f\left(x\right)dx}={\displaystyle \underset{-7}{\overset{5}{\int }}\left(x^2-3x\right)dx}$