Question

what is "In" in definite integral?

Answer 1

It's \[x \ln x -x\] would you like me to prove it?

Answer 2

well, xlnx - x evaluated between your two points

Answer 3

explain a little bit more plz..

Answer 4

can i send u a unsolved file ?

Answer 5

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
\[[ue^u]_{u= \ln a}^{u=\ln b} - \displaystyle\int_{\ln a}^{\ln b}e^udu\]
\[= \left[(u-1)e^u\right]_{u=\ln a}^{u=\ln b}\]
\[=\left[(\ln x -1)x\right]_{x=a}^{x=b}\]
\[=b(\ln b-1) - a(\ln a -1)\]
\[=\ln b^b - \ln a^a +(a-b)\]
\[=\ln \frac{b^b}{a^a} + a -b\]

Answer 6

I can't open it

Answer 7

why?