How should I go about proving the following:
$$\int_1^4 g(\sqrt{x}) dx = 2 \int_1^2 xg(x) dx$$
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$\begingroup$ Let $\sqrt{x}=t$, $\endgroup$– Anurag AOct 15, 2017 at 9:40
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1 Answer
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$$\int_1^4g(\sqrt{x})dx$$ $$x=u^2$$ $$dx=2udu$$ $$\int_1^4g(\sqrt{x})dx=\int_1^22g(\sqrt{u^2})udu=2\int_1^2ug(u)du=2\int_1^2xg(x)dx$$
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$\begingroup$ Aww thanks man! I am so close. Watch somebody downvote this now. $\endgroup$ Oct 15, 2017 at 9:43
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1$\begingroup$ Hah now if somebody downvoted you after reading the previous comments, the system should report that as abuse $\endgroup$– Mr PieOct 15, 2017 at 9:46
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$\begingroup$ @HarryAlli That's what I thought I should do but I don't understand how it turns from $\int_1^4$ to $\int_1^2$ $\endgroup$– JohnDoeOct 15, 2017 at 18:35
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$\begingroup$ @JohnDoe look more deeply into u-substitution $\endgroup$– Mr PieOct 15, 2017 at 21:05
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$\begingroup$ @JohnDoe you have to change the bounds with a u substitution. I suggest you search definite integrals under u-substitution $\endgroup$ Oct 15, 2017 at 22:59