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Core 3 - Integration Help!

Right, I've differentiated part a) which has given me lnx + 1. However I don't know how to use this or where to start with part b), the integration bit.

http://img96.imageshack.us/img96/9764/integrationr.png

Could anyone help me out please?

Much appreciated!
Reply 1
ddx(xlnx)=lnx+1\dfrac{d}{dx}(xlnx)=lnx+1

so, integrating both sides w.r.t. x: ddx(xlnx) dx=lnx dx+1 dx\displaystyle\int\frac{d}{dx}(xlnx)\ dx=\int lnx\ dx+\int 1\ dx

so lnx dx=ddx(xlnx) dx1 dx\displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx
Reply 2
Original post by Pheylan
ddx(xlnx)=lnx+1\dfrac{d}{dx}(xlnx)=lnx+1

so, integrating both sides w.r.t. x: ddx(xlnx) dx=lnx dx+1 dx\displaystyle\int\frac{d}{dx}(xlnx)\ dx=\int lnx\ dx+\int 1\ dx

so lnx dx=ddx(xlnx) dx1 dx\displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx


Sorry, I still don't really understand what is actually going on there. That's pretty much what it says in the mark scheme and I just need it explained really.
Reply 3
Original post by themagicpiano
Sorry, I still don't really understand what is actually going on there. That's pretty much what it says in the mark scheme and I just need it explained really.


well from the first part, you know that ddx(xlnx)=lnx+1\dfrac{d}{dx}(xlnx)=lnx+1

so if you integrate both sides of that equation (which you're allowed to do), you get ddx(xlnx) dx=lnx dx+1 dx\displaystyle\int\frac{d}{dx}(xlnx)\ dx=\int lnx\ dx+\int 1\ dx

subtract 1 dx\displaystyle\int 1\ dx from both sides to get lnx dx\displaystyle\int lnx\ dx on its own:

lnx dx=ddx(xlnx) dx1 dx\displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx

then you can simplify the RHS. (remember that ddx(f(x)) dx=f(x)\displaystyle\int\frac{d}{dx}(f(x))\ dx=f(x) and xn dx=xn+1n+1+c\displaystyle\int x^n\ dx=\frac{x^{n+1}}{n+1}+c )
Reply 4
Original post by Pheylan
well from the first part, you know that ddx(xlnx)=lnx+1\dfrac{d}{dx}(xlnx)=lnx+1

so if you integrate both sides of that equation (which you're allowed to do), you get ddx(xlnx) dx=lnx dx+1 dx\displaystyle\int\frac{d}{dx}(xlnx)\ dx=\int lnx\ dx+\int 1\ dx

subtract 1 dx\displaystyle\int 1\ dx from both sides to get lnx dx\displaystyle\int lnx\ dx on its own:

lnx dx=ddx(xlnx) dx1 dx\displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx

then you can simplify the RHS. (remember that ddx(f(x)) dx=f(x)\displaystyle\int\frac{d}{dx}(f(x))\ dx=f(x) and xn dx=xn+1n+1+c\displaystyle\int x^n\ dx=\frac{x^{n+1}}{n+1}+c )


Ah, that makes sense. Thank you for your help! :smile:
Reply 5
Original post by Pheylan
well from the first part, you know that ddx(xlnx)=lnx+1\dfrac{d}{dx}(xlnx)=lnx+1

so if you integrate both sides of that equation (which you're allowed to do), you get ddx(xlnx) dx=lnx dx+1 dx\displaystyle\int\frac{d}{dx}(xlnx)\ dx=\int lnx\ dx+\int 1\ dx

subtract 1 dx\displaystyle\int 1\ dx from both sides to get lnx dx\displaystyle\int lnx\ dx on its own:

lnx dx=ddx(xlnx) dx1 dx\displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx

then you can simplify the RHS. (remember that ddx(f(x)) dx=f(x)\displaystyle\int\frac{d}{dx}(f(x))\ dx=f(x) and xn dx=xn+1n+1+c\displaystyle\int x^n\ dx=\frac{x^{n+1}}{n+1}+c )


Out of interest, since it says 'Hence, or otherwise find', which method could be used to find the integral if I didn't do part a)?

Thanks again for your help!
Reply 6
Original post by themagicpiano
Out of interest, since it says 'Hence, or otherwise find', which method could be used to find the integral if I didn't do part a)?

have you learnt integration by parts yet? if so:

lnx dx1.lnx dx\displaystyle\int lnx\ dx\equiv\int 1.lnx\ dx

use IBP with u=lnx and dv/dx=1

Original post by themagicpiano
Thanks again for your help!


no problem :smile:

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