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April 16, 2014

April 16, 2014

Posted by **adela** on Sunday, January 2, 2011 at 4:39am.

- calculus -
**MathMate**, Sunday, January 2, 2011 at 8:48amThe first fundamental theorem of calculus states that if f(x) is continuous, real and defined on [a,b], and

F(x)=∫f(x)dx from a to b

then

F(x) is continuous on [a,b] and differentiable on (a,b), then

F'(x) = f(x).

In this case f(t)=sqrt(t)sin(t), definite integral is calculated from x³ to √(x).

Thus if the above theorem to apply, t must be non-negative, which implies that x>0.

If the condition is satisfied, then f(t) is continuous and defined on [0,∞], and consequently F'(t) = f(t).

- calculus -
**Ooooooo**, Friday, October 12, 2012 at 6:43pmYoto

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