[itex]h(x)= \int_0^x (\int_0^uf(t)dt). du[/itex], then why is [itex]h'(x) = \int_0^uf(t)dt[/itex]? Shouldn't it be ##(adsbygoogle = window.adsbygoogle || []).push({});

h(x) - h(0)## in the first equation? where ##h(x)## is the antiderivative of [itex]\int_0^uf(t)dt[/itex]? But wait, isn't antiderivative of a function without limits on it? Like for [itex]\int_a^bf(x)dx[/itex] we would say, let ##F(x)## be the antiderivative of ##f(x)##, i.e. [itex]F(x) = ∫f(x)dx[/itex]. And then we apply limits on ##F(x)## do evaluate the definite integral. So what does ##h(x)## mean in the beginning? Does it mean that ##h(x)## is the antiderivative of [itex]\int_0^uf(t)dt[/itex], i.e. [itex]h(x) = ∫(\int_0^uf(t)dt).dx[/itex] and then we apply the limits 0 and x on it? Would [itex]\int_0^uf(t)dt[/itex] be a separate function and not just some single value?

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# Definite Integral of Definite Integral

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