posted by adela .
using the fundamental theorem of calculus what is the derivative of the definite integral from x^3 to sqrt x of (sqrt t) sin t dt
The 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
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).