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c3 maths help - integration

19397758_1625327037518228_797740212_n.jpgquestion 9 iv (the last thing) the answer to the question before i got as the integral of 1/2te^-t where t = x^2

please help im so stuck
Original post by malaysia007
question 9 iv (the last thing) the answer to the question before i got as the integral of 1/2te^-t where t = x^2

please help im so stuck


Note that you have ∫abf(x).dx=∫a2b2kteβˆ’t.dt\displaystyle \int_a^b f(x) .dx = \int_{a^2}^{b^2}kte^{-t} .dt
Original post by RDKGames
Note that you have ∫abf(x).dx=∫a2b2kteβˆ’t.dt\displaystyle \int_a^b f(x) .dx = \int_{a^2}^{b^2}kte^{-t} .dt


yes so the boundaries would be 4 and 0?
Original post by malaysia007
yes so the boundaries would be 4 and 0?


Yes.
Original post by RDKGames
Yes.


how do i type like with integrals and stuff on here like you haha?
bump
Original post by malaysia007
how do i type like with integrals and stuff on here like you haha?


https://www.thestudentroom.co.uk/help/latex

Original post by malaysia007
bump


Are you still stuck??


thanks haha and yes :/ cant integrate 1/2te^-t between 0 and 4
Original post by malaysia007
thanks haha and yes :/ cant integrate 1/2te^-t between 0 and 4


Just integrate it by parts.
IMG_3660.jpgWhat am i doing wrong
Original post by malaysia007
What am i doing wrong


Getting confused with all the x's t's u's and v's by the looks of it.

Look, we want to evaluate I=∫02x3eβˆ’x2.dx\displaystyle I=\int_0^2 x^3e^{-x^2} .dx

Using the substitution t=x2t=x^2 you turn this integral into I=∫0412teβˆ’t.dt\displaystyle I=\int_0^4 \frac{1}{2}te^{-t}.dt

Now this integral in terms of tt can be evaluated using integration by parts.

Have a go at that. You shouldn't need to even touch xx terms from this point onwards.
Original post by RDKGames
Getting confused with all the x's t's u's and v's by the looks of it.

Look, we want to evaluate I=∫02x3eβˆ’x2.dx\displaystyle I=\int_0^2 x^3e^{-x^2} .dx

Using the substitution t=x2t=x^2 you turn this integral into I=∫0412teβˆ’t.dt\displaystyle I=\int_0^4 \frac{1}{2}te^{-t}.dt

Now this integral in terms of tt can be evaluated using integration by parts.

Have a go at that. You shouldn't need to even touch xx terms from this point onwards.


ok ok thanks, would you advise bringing the 1/2 out
Original post by malaysia007
ok ok thanks, would you advise bringing the 1/2 out


Yes.
I still cant do it its really bugging me
Original post by malaysia007
I still cant do it its really bugging me


Post your working then and we'll see where you fall short on.

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