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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

  • calculus - ,

    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).

  • calculus - ,


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