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
Hello there. <span> 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 </span> 39.
We're looking for the two values being subtracted here. One of these values is easy to find:
<span>g(1) = ∫f(t)dt = 0</span><span> 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:
</span><span>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. </span> I hope my answer has come to your help. God bless and have a nice day ahead!
