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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 A
    Oct 15, 2017 at 9:40

1 Answer 1

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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. $\endgroup$
    – Harry Alli
    Oct 15, 2017 at 9:43
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    $\begingroup$ Hah now if somebody downvoted you after reading the previous comments, the system should report that as abuse $\endgroup$
    – Mr Pie
    Oct 15, 2017 at 9:46
  • $\begingroup$ @HarryAlli That's what I thought I should do but I don't understand how it turns from $\int_1^4$ to $\int_1^2$ $\endgroup$
    – JohnDoe
    Oct 15, 2017 at 18:35
  • $\begingroup$ @JohnDoe look more deeply into u-substitution $\endgroup$
    – Mr Pie
    Oct 15, 2017 at 21:05
  • $\begingroup$ @JohnDoe you have to change the bounds with a u substitution. I suggest you search definite integrals under u-substitution $\endgroup$
    – Harry Alli
    Oct 15, 2017 at 22:59

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