Question

The function f is continuous on the closed interval [1,15] and has the values shown on the table above. Let g(x) = ∫f(t) dt [1,x]. Using the intervals [1,3], [3,6], [6,10], [10,15], what is the approximation of g(15) – g(1) obtained from a left Riemann Sum?
t: 1 3 6 10 15
f(t) 2 3 4 2 -1

Answer 1

We're looking for the two values being subtracted here. One of these values is easy to find:

g(1) = ∫f(t)dt = 0
since taking the integral over an interval of length 0 is 0.
 
The other value we find by taking a Left Riemann Sum, which means that we divide the interval [1,15] into the intervals listed above and find the area of rectangles over those regions:

Each integral breaks down like so:

(3-1)*f(1)=4

(6-3)*f(3)=9

(10-6)*f(6)=16

(15-10)*f(10)=10.

So, the sum of all these integrals is 39, which means g(15)=39.
 
Then, g(15)-g(1)=39-0=39.

I hope my answer has come to your help. God bless and have a nice day ahead!

Answer 2

Hello there.

The function f is continuous on the closed interval [1,15] and has the values shown on the table above. Let g(x) = ∫f(t) dt [1,x]. Using the intervals [1,3], [3,6], [6,10], [10,15], what is the approximation of g(15) – g(1) obtained from a left Riemann Sum?
t: 1 3 6 10 15
f(t) 2 3 4 2 -1

39.