Question

Riemann sums: Calculate each of the following Riemann sum

approximations for the definite integral of f on [a, b], using the

given value of n.

The left sum for $f\left(x\right)=\sqrt{x}$on $\left[0,4\right]$ ,$n=4$.

Short answer

The value is$6·146$.

Step-by-step solution

Step 1. Given information .

Consider the given function$f\left(x\right)=\sqrt{x}$.

Step 2.  Formula used .

The Riemann sum formula for left sum is $L_n=\sum _{i=0}^{n-1}f\left(x_i\right)·\delta x$where $L_n$is the left sum with the n rectangle and$\delta x=\frac{b-a}{n}$.

Step 3. Find the value .

Divided the open interval $\left[0,4\right]$in 4 sub intervals $\left(0,1\right),\left(1,2\right),\left(2,3\right),\left(3,4\right)$where $x_0=0,x_1=1,x_2=2,x_3=3,x_4=4$

$\delta x=\frac{b-a}{h}=\frac{4-0}{4}=1$

Substitute the value of $\delta x$and $x_0........x_4$in the given formula .

$L_n=\sum _{i=0}^{4}f\left(x_i\right)·\delta x$

$$\begin{gathered} L_n=f\left(x_0\right)·\delta x+f\left(x_1\right)·\delta x+f\left(x_2\right)·\delta x+f\left(x_3\right)·\delta x_3+f\left(x_4\right)·\delta x \\ =0\times 1+1\times 1+\sqrt{2}\times 1+\sqrt{3}\times 1+2\times 1 \\ =3+\sqrt{2}+\sqrt{3} \\ =6·146 \end{gathered}$$