The Fundamental Theorem of Calculus

Question

f(x) = ∫ [x^2, 1] t^5 dt

f ' (x) =

Answer 1

could say

f(x)=-(t⁶)/6 from t=1 to x²,,,,just integrating -∫t⁵dt bound 1 to x²

f(x)=-[(x¹²)/6]+1/6

f'(x)=-2(x¹¹)

Answer 2

Hi,

f(x) = ∫ [x^2, 1] t^5 dt = - ∫ [1, x²] t^5 dt First, change the bounded of integral

f ' (x) = - (x²)' * (x²)⁵

f ' (x) = - (2x) * (x²)⁵

f ' (x) = - 2 x¹¹

^{I hope it is useful,}

^Minoo