integration by parts

Hi,

I have to integrate this by parts:

(x-5)(e^(-x)) and it says you have to treat (x-5) as dv/dx and exp(-x) as u.

This is a past exam paper and so I have the solution, however I get a different answer. I get:

\displaystyle \int_a^b \{(x-5) exp(-x)\} dx = \left[ [1/2]x^2 -cx \right]_a^b

As the first part but I should apparently get:

\displaystyle \int_a^b \{(x-5) exp(-x)\} dx = \left[ [1/2] (x-c)^2 \right]_a^b

Can anyone see what is wrong?

thanks

Reply 1

The_Right_Ordinary

They're the same.
i don't know how to use latex, but i'll try to explain

The (x-c) expands to x^2 - 2cx + c^2. When divided by two this gives (1/2)x^2 - cx and the c^2 will cancel out when you substitute in a and b.

Therefore it's equal to the answer given.

Reply 2

Big_Sam

OP

ok i understand that bit, but in exam how do i know which way to integrate it i.e my way or that way

Reply 3

The_Right_Ordinary

The examiners know that there's more than one way to write the same thing so they'd mark them both correct and it wouldn't matter

Your method is perfectly fine as long as you can manage to consistently get the right answers.

Reply 4

steve2005

17

Big_Sam

Hi,

I have to integrate this by parts:

(x-5)(e^(-x)) and it says you have to treat (x-5) as dv/dx and exp(-x) as u.

This is a past exam paper and so I have the solution, however I get a different answer. I get:

\displaystyle \int_a^b \{(x-5) exp(-x)\} dx = \left[ [1/2]x^2 -cx \right]_a^b

As the first part but I should apparently get:

\displaystyle \int_a^b \{(x-5) exp(-x)\} dx = \left[ [1/2] (x-c)^2 \right]_a^b

Can anyone see what is wrong?

thanks

When you integrate

(x-5)e^{-x}

you get,according to Wolfram

e^{-x}(4-x)

so I am puzzled by the question. It does not make sense to me.

Reply 5

astudent

7

steve2005

When you integrate

(x-5)e^{-x}

you get,according to Wolfram

e^{-x}(4-x)

so I am puzzled by the question. It does not make sense to me.

Ive just integrated it and putting in the limits is exactly what u get above.

Im confused as to where the C came from and where the e^-x disappeared in the op's answer.

EDIT: also whats the point of integrating (x-5), surely it makes much more sense to differentiate it ?

Reply 6

steve2005

17

astudent

Ive just integrated it and putting in the limits is exactly what u get above.

Im confused as to where the C came from and where the e^-x disappeared in the op's answer.

EDIT: also whats the point of integrating (x-5), surely it makes much more sense to differentiate it ?