There are two theorems:(adsbygoogle = window.adsbygoogle || []).push({});

The fundamental theorem of calculus: [tex]\int_{a}^{b}F'(x) = F(b) - F(a)[/tex]

And the theorem that states if f is continious on [a,b]and g:[a,b]->R is defined by

g(x) = [tex]\int_{a}^{x}f(t) dt[/tex], then g is differentiable on (a,b) and;

g'(x) = [tex]\frac{d}{dx}(\int^{x}_{a}f(t)) = f(x)[/tex]

"There’s one question that asks if it is possible to readily show that each one implies the other, by a few lines of manipulation."

Does anyone know how to do that?

Let f be cts on [a,b], and let F(x) = [tex]\int f(x) dx[/tex] be called F, the antiderivative of f. Then

[tex]\int_{a}^{b}f(x) dx = F(b) - F(a)[/tex]

By theorem [tex]\frac{d}{dx}(\int^{x}_{a}f(t)) = f(x)[/tex], we know [tex]\int_{a}^{x}f(t) dt = F(b) - F(a)[/tex] & F(x) has the same derivative, f(x). Hence there will be a constant c such that [tex]\int_{a}^{x}f(t)dt = F(x)+c[/tex] When x = a, we get

F(a) + c = [tex]\int_{a}^{a}f(t)dt = 0[/tex] so c = -F(a)

Hence, [tex]\int_{a}^{x}f(t)dt [/tex] = F(x)-F(a)

When x = b this yields

[tex]\int_{a}^{b}f(t)dt = F(b)-F(a)[/tex]

I'm not sure if this really answers the question though.

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# Fundamental theorem of calculus

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