if there is an integral (integral(from 0 to 8) T'(X)dx),
does this simply mean the Y value when T(X) is 8?
2007-04-22 14:28:58 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You kind of have the right idea, but...
According to one of the Fundamental Theorem's of Calculus, it means T(8) - T(0). If the function had a value of 0 at T(0), then you obviously wouldn't have to include the T(0).
2007-04-22 14:31:36 · answer #1 · answered by عبد الله (ドラゴン) 5 · 0⤊ 2⤋
No. Use the fundamental theorem of calculus.
It means T(8) - T(0).
The answer depends upon T(0).
2007-04-22 21:34:44 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
No. This is known as a "closed integral" since it has real limits at both ends. T(X) is evaluated at the upper end and the lower end, and closed integral is T(X) at X=8 - T(X) at X=0.
2007-04-22 21:32:55 · answer #3 · answered by cattbarf 7 · 0⤊ 1⤋
You're close, it would be T(8) - T(0)
2007-04-22 21:31:49 · answer #4 · answered by Kathleen K 7 · 0⤊ 1⤋