what is "In" in definite integral?

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what is "In" in definite integral?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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It's \[x \ln x -x\] would you like me to prove it?
well, xlnx - x evaluated between your two points
explain a little bit more plz..

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can i send u a unsolved file ?
Let's prove it then... \[\displaystyle\int_{a}^{b}\ln x dx.\] Let u=lnx so that dx=xdu, x=e^u. Then our integral becomes \[\displaystyle\int_{\ln a}^{\ln b}ue^udu\] now use integration by parts \[[ue^u]_{u= \ln a}^{u=\ln b} - \displaystyle\int_{\ln a}^{\ln b}e^udu\] \[= \left[(u-1)e^u\right]_{u=\ln a}^{u=\ln b}\] \[=\left[(\ln x -1)x\right]_{x=a}^{x=b}\] \[=b(\ln b-1) - a(\ln a -1)\] \[=\ln b^b - \ln a^a +(a-b)\] \[=\ln \frac{b^b}{a^a} + a -b\]
I can't open it
why?

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