Integration by Substitution

[Laser]

Need your help, please, turk, exam on Wednesday!





In both cases I'm referring to the integration by substitution question.

In very simple terms, what do I have to do?

[Max Rambone]

http://lmgtfy.com/?q=integration+by+substitution

[Laser]

http://lmgtfy.com/?q=integration+by+substitution

no I need someone who I can actually feedback to, I've even tried A-level tutor websites and they are bloody useless.

[Turky]

Need your help, please, turk, exam on Wednesday!





In both cases I'm referring to the integration by substitution question.

In very simple terms, what do I have to do?

sigh I keep telling you to ask me on vent, it's easier...

1. i) simple derivative I hope

ii) You have to recognize that when the numerator is a pseudo-derivative of the denominator (ie, only a constant factor is different), that you then have a natural log integration. (that's the short and sweet version)

The long version:

set u = 2x^2 - 8x + 3, so du/dx = 4x - 8 = 4(x-2). Rearranging the latter, du/4 = (x-2)dx. I did this so that I could plug something directly into the integral in terms of du, since I know that I have an x - 2 term and a dx in the integrand.

So just replace du/4 for (x - 2)dx, and replace 2x^2 - 8x + 3 with u, so your integral becomes (1/4) * int(du/u). I pulled the 1/4 out of the integral to simplify things. Now this becomes a simple integration, but now you must remember to change your limits of integration in terms of u. To do this, simply plug in the limits of integration (4 and 6), and plug them into your u-equation above.

You should find that your new limits of integration are 3 and 27. So we want to integrate (1/4)*du/u from 3 to 27, which means evaluating (1/4)*ln( |u| ) from 3 to 27. So that becomes (1/4) * (ln(27) - ln(3)). To simplify that, recall that ln(a) - ln(b) = ln(a/b). So your final answer for the integral is (1/4) * ln(9) = (1/4)*ln(3^2) = (1/2)*ln(3). (Recall that ln(a^x) = x*ln(a) for that last part.)


ii) They tell you which u to yews in this problem. This means du = 3*dx, dx = du/3. However, this doesn't do anything for that x out front. To fix this, we can solve our u equation for x, so x = (u-1)/3. Now we can plug in for x, sqrt(3x-1), and dx, all directly in terms of u, and that takes care of the whole integral. So that leaves us with (1/9)*int( (u-1)*u^.5) = (1/9) * int(u^1.5 - u^.5). Hopefully that becomes a simple integration for you (it's just inverse power rule).

[Laser]

tyvm <3

[piipe]

turk u are a nerd