My brain melted and I am having trouble seeing why these are equal ($f$ is continuous).
$$\int_0^1 f(x)\,dx=\int_0^1 f(x^2)2x\,dx$$
Thanks in advance.
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Have you ever started with an integral like $\int x\sin(x^2),dx$ and found the antiderivative by substituting $u=x^2$? – Ted Shifrin Jul 08 '16 at 23:01
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no, I guess I skipped that part, bad mistake. – Asinomás Jul 08 '16 at 23:02
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It's just the chain rule. If $F'(u)=f(u)$, then $(F\circ g)'(x) = f(g(x))g'(x)$. Apply that with $g(x)=x^2$ and $F(u) = -\frac12\cos u$. – Ted Shifrin Jul 08 '16 at 23:06
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oh yeah, thanks, that helped, for some reason I got really confused with the integral sign. – Asinomás Jul 08 '16 at 23:10
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Enforce the substitution $x \mapsto x^2$:
$$\int_0^1 f(x)\, dx=\int_{u=0}^{u=1} f(u) \, du=\int_{x^2=0}^{x^2=1} f(x^2) d(x^2) = \int_{x=0}^{x=1} f(x^2) \, d(x^2)=\int_0^1 f(x^2) 2x \, dx$$
As
$$\frac{d(x^2)}{dx}=2x$$
Or if you prefer let $x=u^2$, $\frac{dx}{du}=2u$, $dx=2u \, du$.
Ahmed S. Attaalla
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Think of $x^2$ as one variable $u$, i.e. $u=x^2$ that's why I've put that step there. At least that's how I see it. Maybe what your looking for is a explanation onto why integration by substitution works? @CarryonSmiling – Ahmed S. Attaalla Jul 08 '16 at 22:52
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I think that's what I need, cause I'm not really sure what's going on, thanks btw. – Asinomás Jul 08 '16 at 23:00