integration question (1 Viewer)

[jojo321]

hey everyone,

can someone please help me with this question??

_{1 -}|^{- e} 2lnx.dx

(the _-|^- is the integration sign)

thanks

 

[CM_Tutor]

This problem cannot be done as it stands by 3 unit methods. However, it can be transformed to a valid 3u problem, in at least two ways:

Method One: Differentiate the function xln x, and then integtate the result to establish what int ln x dx is.

Method Two: Draw a diagram of y= 2ln x, and mark the points A(1, 0) - the x-intercept of y = 2ln x - B(e, 0),
C(e, 2) - which lies on the curve - and D(0, 2). The required area is ABC. However, the area OACD can be found by integration against the y-axis (O is the origin), and OBCD is a rectangle of area 2e. So, the required area is
ABC = 2e - area OACD.

Method Three (Extn 2 students only): Use integration by parts.

 

[jojo321]

how would u do it using method 1?? i dont think i understand...

if its not too much trouble would you be able to type it out for me??

thanks

 

[SarahMary]

since it is 3 unit you can also use the substitution method, you know let u=ln x, find du, and do it that way

 

[freaking_out]

Originally posted by jojo321
how would u do it using method 1?? i dont think i understand...

if its not too much trouble would you be able to type it out for me??

thanks

well when u differentiate xlnx, u get an xlnx= lnx + 1

hence when u integrate lnx + 1, it should equal to xlnx...so work from there.

 

[jojo321]

Originally posted by freaking_out
well when u differentiate xlnx, u get an xlnx= lnx + 1

hence when u integrate lnx + 1, it should equal to xlnx...so work from there.

yeh i got that far...but what now? thats the bit im stuck at

 

[George W. Bush]

i.e. xlnx = integral (lnx) + integral (1)
integral (lnx) = xlnx - integral (1)
= xlnx - x + c

and substitution doesn't work because you have no 1/x for du

 

[CM_Tutor]

So, the full solution by method 1 would be:

d/dx (xln x) = ln x * d/dx(x) + x * d/dx(ln x), using the product rule
= ln x * 1 + x * 1 / x
= ln x + 1

Integrating both sides with respect to x, we get:
xln x = int ln x dx + int 1 dx
xln x = int ln x dx + x + C, for some constant C.

So, int ln x dx = xln x - x + C', where C' = -C

Now, we seek int (from 1 to e) 2ln x dx = 2 * int (from 1 to e) ln x dx
= 2 * [xln x - x] from 1 to e
= 2 * [(e * ln e - e) - (1 * ln 1 - 1)]
= 2 * [e * 1 - e - 1 * 0 + 1]
= 2

And GWB, the substitution u = ln x can (and does ) work, but I wouldn't encourage a 3u student to try it.

 

[KeypadSDM]

Let u = Ln[x]
x = e^u

e^udu = dx

:.
/
| 2ln[x]dx
/
=
/
|2ue^udu
/

Note, d/du ue^u = e^u + ue^u

:. ue^u = d/du ue^u - e^u
/
|2ue^udu
/
= 2ue^u -
/
|2e^udu
/
=2ue^u - 2e^u + C

I think ...

 

[George W. Bush]

my apologies, it does indeed. I wasn't thinking laterally enough