Question

consider the integral (int_{1}^{4}(2x^{2}+4x + 1)dx). (a) find the riemann sum for this integral using right endpoints and (n = 3). (b) find the riemann sum for this same integral, using left endpoints and (n = 3). video example: solving a similar problem

Explanation:

Step1: Determine the width of sub - intervals

The integral is $\int_{1}^{4}(2x^{2}+4x + 1)dx$ with $n = 3$. The width of each sub - interval $\Delta x=\frac{b - a}{n}$, where $a = 1$ and $b = 4$. So, $\Delta x=\frac{4 - 1}{3}=1$.

Step2: Find the right - endpoint Riemann sum ($R_3$)

The right - endpoints of the sub - intervals $[1,2]$, $[2,3]$, $[3,4]$ are $x_1 = 2$, $x_2=3$, $x_3 = 4$.
The function is $f(x)=2x^{2}+4x + 1$.
$f(x_1)=2(2)^{2}+4(2)+1=8 + 8+1 = 17$.
$f(x_2)=2(3)^{2}+4(3)+1=18 + 12 + 1=31$.
$f(x_3)=2(4)^{2}+4(4)+1=32+16 + 1=49$.
$R_3=\sum_{i = 1}^{3}f(x_i)\Delta x=(17 + 31+49)\times1=97$.

Step3: Find the left - endpoint Riemann sum ($L_3$)

The left - endpoints of the sub - intervals $[1,2]$, $[2,3]$, $[3,4]$ are $x_0 = 1$, $x_1=2$, $x_2 = 3$.
$f(x_0)=2(1)^{2}+4(1)+1=2 + 4+1 = 7$.
$f(x_1)=2(2)^{2}+4(2)+1=8 + 8+1 = 17$.
$f(x_2)=2(3)^{2}+4(3)+1=18 + 12 + 1=31$.
$L_3=\sum_{i = 0}^{2}f(x_i)\Delta x=(7 + 17+31)\times1=55$.

Answer:

$R_3 = 97$, $L_3=55$