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: 39

Answer 1

We're looking for the two values being subtracted here. One of these values is easy to find:
\[g(1)=\int\limits_{1}^{1}f(t)dt=0\] 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:
\[g(15)=\int\limits_{1}^{15}f(t)dt = \int\limits_{1}^{3}f(t)dt + \int\limits_{3}^{6}f(t)dt + \int\limits_{6}^{10}f(t)dt + \int\limits_{10}^{15}f(t)dt\]
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.

Answer 2

thank you so much!:)