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This is the original definite integral. $$\int_1^3\frac{2x-3}{\sqrt{4x-x^2}}$$ I do not know what to do after completing the square and I am stuck at that specific part after plugging back in. $$(4x-x^2)$$ $$-(x^2-4x )$$ $$-(x^2-4x+4)-4$$ $$-(x-2)^2-4$$ So I would plug that in back to the integral $$\int_1^3\frac{2x-3}{\sqrt{-(x-2)^2-4}}$$ and this is where I am not sure what to do next I was thinking u-substitution but what would be my u?

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  • $\begingroup$ First, need to add dx's to your integrals. Second, have you thought about making a trig substitution? $\endgroup$ – David Sep 20 '15 at 21:48
  • $\begingroup$ Hint First, note that you have a mistake since $4x-x^2=4-(x-2)^2$. Second, use that $$ \int \frac{du}{\sqrt{1-u^2}}=\sin^{-1}(u)+C,$$ in your case you will have to split the integral and use $u=(x-2)/2$ $\endgroup$ – Alonso Delfín Sep 20 '15 at 22:02
  • $\begingroup$ I have another question lets say i was taking the integral of another equation which required me to complete the square and these were the values $x^4+x^2+C$ when i factor out to complete the square would I also factor out a $x^2$? leading it to this $x^2(x^2+1)+C$ ? $\endgroup$ – Carlos V Sep 21 '15 at 1:07
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Oops, I think you made a small algebra slip (no worries, it happens to the best of us!). You need to swap the -4 for +4 in your working since:

$-(x^2-4x) = -(x^2-4x)-4+4 = -(x^2-4x+4)+4 = 4-(x-2)^2$

As for a hint for the rest of it, you could notice that:

$\frac{d} {dx} (4-(x-2)^2) = 4-2x$

Using this you could split you integral into 2 parts by writing the numerator of your fraction as: $ -(4-2x) + 1 $
This would leave two integrals one which you can find an anti derivative for and another which you can use a trig substitution to solve. (Try thinking about $cos^2x + sin^2x = 1$ if you're stuck!) Hope this helps.

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  • $\begingroup$ Would you mind elaborating on why it would be +4 instead of -4 on the outside i am a bit confused on that. $\endgroup$ – Carlos V Sep 21 '15 at 0:57
  • $\begingroup$ Oh never mind, another user pointed it out thank you anyways. $\endgroup$ – Carlos V Sep 21 '15 at 1:04

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