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Core 3 Substitution by parts Q

See attached.Jan 07 4B integrate by sub.png

I'm a bit confused.

I get INTEGRAL (SQRT(U-5) (U^1\2) du/2


But what next?

The mark scheme doesnt help.. Apparently, the x at the start of the question, dissapears somehow but when i did the q, I put that the x becomes (SQRT(U-5)

Thanks in advance :smile:
Reply 1
Original post by MSI_10
See attached.Jan 07 4B integrate by sub.png

I'm a bit confused.

I get INTEGRAL (SQRT(U-5) (U^1\2) du/2


But what next?

The mark scheme doesnt help.. Apparently, the x at the start of the question, dissapears somehow but when i did the q, I put that the x becomes (SQRT(U-5)

Thanks in advance :smile:


u=x2+5u=x^2+5

dudx=2x\frac{du}{dx}=2x

dx=du2xdx=\frac{du}{2x}

x×u12du2x\int x \times u^{\frac{1}{2}} \frac{du}{2x}

Can you see that the x will cancel out now?

So becomes

12u12du\frac{1}{2} \int u^{\frac{1}{2}} du

=12×23u32+C=\frac{1}{2} \times \frac{2}{3}u^{\frac{3}{2}}+C

=13(x2+5)32+C=\frac{1}{3}(x^2+5)^{\frac{3}{2}}+C
(edited 11 years ago)
Reply 2
You need to work out du/dx :smile:
Reply 3
Original post by Future Me
u=x2+5u=x^2+5

dudx=2x\frac{du}{dx}=2x

dx=du2xdx=\frac{du}{2x}

x×u12du2x\int x \times u^{\frac{1}{2}} \frac{du}{2x}

Can you see that the x will cancel out now?

So becomes

12u12du\frac{1}{2} \int u^{\frac{1}{2}} du

=12×23u32+C=\frac{1}{2} \times \frac{2}{3}u^{\frac{3}{2}}+C

=13(x2+5)32+C=\frac{1}{3}(x^2+5)^{\frac{3}{2}}+C


OHHHHHHHH crap I forgot about that! :P

thanks I understand now :smile:

Original post by jaheen22
You need to work out du/dx :smile:



YeAH I did i got du/2 but forgot it was supposed to be 2x not 2 :P

thanks all :smile:

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