Chapter 5 - The Integral - 5.1 Approximating and Computing Area - Exercises - Page 235: 17

$$87$$

Given $$ f(x)=4 x+3, \quad[5,8]$$ Since $n= 6$, $\Delta x=\dfrac{1}{2}$ and $$ x_0= 5,\ x_1=5.5,\ x_2=6,\ x_3=6.5,\ x_4=7,\ x_5=7.5,\ x_6=8$$ Then \begin{align*} M_n&=\left[f\left(\frac{x_{0}+x_{1}}{2}\right)+\cdots+f\left(\frac{x_{N-1}+x_{N}}{2}\right)\right] \Delta x\\ M_6&= \left[f(5.25)+f(5.75)+f(6.25)+f(6.75)+f(7.25)+f(7.75) \right] \Delta x\\ &= [ 24+26+28+30+32+34]\frac{1}{2}\\ &=87 \end{align*}