Answer
$$L_{4}=3.875, R_{4}=4.375, M_{4}=4.425 $$
Work Step by Step
Since $n=4$, on $[1,3]$ $\Delta x= \dfrac{b-a}{n}=\dfrac{1}{2}$ and $$x_0= 1,\ x_1= 1.5,\ x_2= 2 ,\ x_3=2.5,\ x_4= 3 $$ Then \begin{align*} L_{n}&=\left[f(x_0)+f(x_1)+.......+f(x_{n-1})\right]\Delta x\\ L_4&=\left[f(x_0)+f(x_1)+.......+f(x_{3})\right]\Delta x\\ &=\left[ f(1)+ f( 1.5)+ f( 2)+f( 2.5) \right]\frac{1}{2},\ \ \text{from the given figure }\\ &= \left[ 1+2+2.5+2.25\right]\frac{1}{2}\\ &= 3.875 \end{align*} To find $R_4$ \begin{align*} R_{n}&=\left[ f(x_1)+.......+f(x_{n})\right]\Delta x\\ R_4&=\left[ f(x_1)+.......+f(x_{4})\right]\Delta x\\ &=\left[ f( 1.5)+ f( 2)+f( 2.5)+f(3) \right]\frac{1}{2},\ \ \text{from the given figure }\\ &= \left[ 2+2.5+2.25+2\right]\frac{1}{2}\\ &= 4.375 \end{align*} To find $M_4$ \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_4&= \left[f(1.25)+f(1.75)+f(2.25)+f(2.75) \right] \Delta x,\ \ \text{from the given figure }\\ &= [ 1.5+2.25+2.9+2.2]\frac{1}{2}\\ &=4.425
\end{align*}
