Question

use the concept of the detinite integral to find the total area between the graph off(x) and the x-axis by taking the limit of the associated right Riemann sum. Write the exact answer. Do not round. (Hint: Extra care is needed on those intervals where< 0. Remember that the definite integral represents a signed area.)

Answer 1

Given:

$\int ^6_02x+3$

To find the area between the graph of f(x) and the x-axis by taking the limit of the associated right Riemann sum:

Here, a=0, b=6 and f(x)=2x+3

Let the number of rectangles is, n=6.

The formula for the right riemann sum is,

$\int ^b_af(x)dx=\Delta x(f\mleft(x_0\mright)+f(x_1)+f(x_2)+...+f(x_{n-2})+f(x_{n-1})$

Here,

$\begin{gathered} \Delta x=\frac{b-a}{n} \\ =\frac{6-0}{6} \\ =1 \end{gathered}$

Divide the intervals [0,6] into n=6 subintervals with the length Δx=1 for the following endpoints. we get,

0, 1, 2, 3, 4, 5, 6.

Since, using the right riemann sum,

f(1)=2(1)+3

f(1)=5

f(2)=2(2)+3

f(2)=7

f(3)=2(3)+3

f(3)=9

f(4)=2(4)+3

f(4)=11

f(5)=2(5)+3

f(5)=13

f(6)=2(6)+3

f(6)=15

Hence, the area is,

$\begin{gathered} \int ^6_02x+3=1(5+7+9+11+13+15) \\ =60\text{ square units.} \end{gathered}$

Hence, the area between the graph of f(x) and the x-axis is 60 square units.