Integration

I am struggling to find the value of the integral of x^2/(4-x) between the limits 9 and 0. I have tried using substitution but I then end up with a negative number in lnx which can't be right :/ Help please.

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Reply 1

Europa192

Oops, I used the wrong bounds - no wonder it wasn't working.

Reply 2

Europa192

Ok, the integral is now between 3 and 0 but I am still getting the wrong answer

Reply 3

joostan

Original post by Europa192

Ok, the integral is now between 3 and 0 but I am still getting the wrong answer

What have you tried? - If you post some working I can find where you went wrong. :smile:

Reply 4

nmanvi

Original post by Europa192

hmmm try doing integration by parts with x^2 x (4-x)^-1

Reply 5

ghostwalker

Original post by Europa192

Ok, the integral is now between 3 and 0 but I am still getting the wrong answer

Either way your bounds seem a little strange. Can you post the actual question?

Reply 6

joostan

Original post by nmanvi

hmmm try doing integration by parts with x^2 x (4-x)^-1

I'd have thought partial fractions or even a substitution would have been simpler than that . . . :tongue:

Reply 7

nmanvi

Original post by joostan

I'd have thought partial fractions or even a substitution would have been simpler than that . . . :tongue:

soz havent done C4 in like over a month now XD
but how can you do partial fractions if there is only one bracket as a denominator?

Reply 8

joostan

Original post by nmanvi

soz havent done C4 in like over a month now XD
but how can you do partial fractions if there is only one bracket as a denominator?

Spoiler

Reply 9

nmanvi

Original post by joostan

Spoiler

im currently working it out and substitution looks much easier :tongue:

yah

Reply 10

nmanvi

working (tell if wrong):

u = 4-x

x = 4 - u

x^2 = u^2 -8u + 16

when x = 3 , u = 1 // when x = 0 , u = 4 so limits are 1 and 4 (not exact sure)

integrate:

(u^2 - 8u +16)/u

u -8 +16/u

(after integration) =

u^2/2 -8u +16ln|u|

continue with the limits 1 and 4 and tell me how it works out

(edited 10 years ago)

Reply 11

joostan

Original post by nmanvi

Spoiler

(edited 10 years ago)

Reply 12

Jkn

Original post by Europa192

The most direct method (that I find rather satisfying) would be to simply use algebraic manipulation:

Spoiler

Reply 13

TenOfThem

You could do long division with the original and then integrate

Reply 14

joostan

Original post by Jkn

The most direct method (that I find rather satisfying) would be to simply use algebraic manipulation:

Spoiler

My preferred method too, it's so much tidier :tongue:

Reply 15

Jkn

Original post by joostan

My preferred method too, it's so much tidier :tongue:

I don't know why people are keen on it? An even nicer trick would be to note that

\displaystyle x^2=((x-4)+4)^2=(x-4)^2+8(x-4)+16

Reply 16

joostan

Original post by Jkn

I don't know why people are keen on it? An even nicer trick would be to note that

\displaystyle x^2=((x-4)+4)^2=(x-4)^2+8(x-4)+16

Neat, but that's essentially the same thing . .. but everyone prefers to go down the boring methodical partial fractions route or the lengthy algebraic long division.
Anyhow, we seem to be clogging up the OPs thread with matters of little relevance.

Reply 17

Jkn

Original post by joostan

This is the problem. Whilst it is 'essentially the same thing' you must appreciate that, when generalised, the two methods are entirely different. For example, express

f(x)

in the form

f(x-a)

where f is a polynomial. The first method would likely be useless as it is a special case, though the second may be more useful. For this particular problem, there is another neat little trick that could be used. I wonder if you know what I am referring to? :colone:

Reply 18

joostan

Original post by Jkn

This is the problem. Whilst it is 'essentially the same thing' you must appreciate that, when generalised, the two methods are entirely different. For example, express

f(x)

in the form

f(x-a)

Well it has similarities with completing the square, but also to the reversal of limits that we discussed yesterday, but I don't know if I know of a particular trick that you are thinking of :confused: