• You have G(x) = integral of 1 to x of f(t) dt where f(t) is some function of t. You can integrate f(t) indefinitely to make F(t) + C all this means is to find some F(t) whose derivative F'(t) is f(t) Then integral of 1 to x f(t) dt is then defined as F(x) - F(1) If G(x) = F(x) - F(1) then G'(x) = F'(x) because F(1) is constant. And F'(x) is f(x) because that is how F is defined. So now run through the argument with your own f(t)

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