Answer:
[tex]A_L=1.75[/tex]
Step-by-step explanation:
We are given:
[tex]f(x)=x^2[/tex]
[tex]interval = [a,b] = [0,2][/tex]
Since [tex]n = 4[/tex] ⇒ [tex]\Delta x = \frac{b-a}{n} = \frac{2-0}{4}=\frac{1}{2}[/tex]
Riemann sum is area under the function given. And it is asked to find Riemann sum for the left endpoint.
[tex]A_L= \sum\limits^{n}_{i=1}\Delta xf(x_i) = \frac{1}{2}(0^2+(\frac{1}{2})^2+1^2+(\frac{3}{2})^2)=\frac{7}{4}=1.75[/tex]
Note:
If it will be asked to find right endpoint too,
[tex]A_R=\sum\limits^{n}_{i=1}\Delta xf(x_i) =\frac{1}{2}((\frac{1}{2})^2+1^2+(\frac{3}{2})^2+2^2)=\frac{15}{4}=3.75[/tex]
The average of left and right endpoint Riemann sums will give approximate result of the area under [tex]f(x)=x^2[/tex] and it can be compared with the result of integral of the same function in the interval given.
So, [tex](A_R+A_L)/2 = (1.75+3.75)/2=2.25[/tex]
[tex]\int^2_0x^2dx=x^3/3|^2_0=8/3=2.67[/tex]
Result are close but not same, since one is approximate and one is exact; however, by increasing sample rates (subintervals), closer result to the exact value can be found.
