Assume s,v, and a are continuous functions and s is twice differentiable, where s(t) represents the position, v(t) is velocity, a(t) is acceleration

Question

 Find the following:

 

d/dx of integral from t to t^2 of v(w)dw

 

integral from t to t^2 of v(w)dw

 

second derivative of integral from t to t^2 of v(w)dw 

 

Calculus question I am stuck on

Answer 1

∫_{(from t to t²)} v(w)dw = ∫_{(from t to 0)}v(w)dw + ∫_{(from 0 to t² )} v(w)dw

 

                              = -∫_{(from 0 to t)}v(w)dw + ∫_{(from 0 to t²)} v(w)dw

 

By the Fundamental Theorem of Calculus, d/dt[∫_{(from t to t²)} v(w)dw] = -v(t) + 2tv(t²)

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∫_{(from t to t²)} v(w)dw = ∫_{(from t to t²)} s'(w)dw = s(w)_{(from t to t²)} = s(t²) - s(t)

 

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d²/dt²[∫_{(from t to t² )}v(w)dw] = d/dt[-v(t) + 2tv(t²)]

 

                                          = -v'(t) + 2v(t²) + 2tv'(t²)(2t)

 

                                          = -a(t) + 2v(t²) + 4t²a(t²)