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Edexcel C3,C4 June 2013 Thread

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Reply 5080
Avatar for orange94
orange94
Original post by OneMorePanda
Good luck integrating ex2e^{x^2}.


Actually I would make my u= e^x^2

And dv= 6x


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Reply 5081
Avatar for Rayquaza
Rayquaza
Original post by justinawe
y=ex2y = e^{x^2}

ln⁡y=ln⁡ex2\ln y = \ln e^{x^2}

ln⁡y=x2\ln y = x^2

Differentiating both sides w.r.t x:

1y×dydx=2x\dfrac{1}{y} \times \dfrac{dy}{dx} = 2x

dydx=2xy=2xex2\dfrac{dy}{dx} = 2xy = 2xe^{x^2}


Does this help you figure out how to integrate that?


Sorry, no...Still lost. Like a cow on Astroturf.
Reply 5082
Avatar for Story
Story
Original post by nm786
no don't expand it.


How come? Is that a general rule for differential equations?
Original post by nm786
sin2xcos2x=(12sin2x)2sin^2xcos^2x =( \frac{1}{2}sin2x )^2 now expand the brackets and integrate.

Hmm interesting i would tackle this q in a different much longer method lol im skacking in my speed of recognizing identities:/
Reply 5084
Avatar for Kieran?
Kieran?
2
Original post by nm786
sin2xcos2x=(12sin2x)2sin^2xcos^2x =( \frac{1}{2}sin2x )^2 now expand the brackets and integrate.


What if it was 1/sin^2(x)cos^2(x) ?


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Original post by justinawe
y=ex2y = e^{x^2}

ln⁡y=ln⁡ex2\ln y = \ln e^{x^2}

ln⁡y=x2\ln y = x^2

Differentiating both sides w.r.t x:

1y×dydx=2x\dfrac{1}{y} \times \dfrac{dy}{dx} = 2x

dydx=2xy=2xex2\dfrac{dy}{dx} = 2xy = 2xe^{x^2}


Does this help you figure out how to integrate that?



can you not use the integration by parts straight away?
Did anyone get to the point where there was not enough time to attempt every single question on every single edexcel paper from several years ago and every soloman paper, so decided to just do the hard questions instead?

I.e. leave out implicit diff and binomial and partial fractions and anything else that looked doable enough
Original post by SortYourLife
Isn't that just (1/2x) e^x^2 ? :s


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No, you could differentiate that to see that it isn't.
Original post by OneMorePanda
Good luck integrating ex2e^{x^2}.


Lol im so undercooked atm how do you integrate that :/
Original post by OneMorePanda
No, you could differentiate that to see that it isn't.


Oh yeah :') you'd just use substitution wouldn't you?


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∫xx−2 dx\displaystyle \int \frac{x}{x-2} \ dx

QQ how would you split the above into partial fractions?
Reply 5091
Avatar for Tikara
Tikara
Original post by überambitious_ox
Did anyone get to the point where there was not enough time to attempt every single question on every single edexcel paper from several years ago and every soloman paper, so decided to just do the hard questions instead?

I.e. leave out implicit diff and binomial and partial fractions and anything else that looked doable enough


exactly what I'm doing lmao - although I'm mainly just focusing on the big integration questions :P
when u get to the point where u have (0.5sin(2x))^2 u just use the double angle formula of cosine and then u integrate
Reply 5093
Avatar for orange94
orange94
Original post by Rayquaza
Book shows to integrate by substitution but I'm lost.


What don't you understand du=2x dx





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Reply 5094
Avatar for Tikara
Tikara
If you times an integral (with limits) by -1 do you swap the limits?
Original post by SortYourLife
Oh yeah :') you'd just use substitution wouldn't you?


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Yeah.

Original post by Myocardium
∫xx−2 dx\displaystyle \int \frac{x}{x-2} \ dx

QQ how would you split the above into partial fractions?


I'm not sure if it's called partial fractions, but you can write the x in the numerator as x - 2 + 2.

Original post by Tikara
If you times an integral (with limits) by -1 do you swap the limits?


Yeah.
Original post by Myocardium
∫xx−2 dx\displaystyle \int \frac{x}{x-2} \ dx

QQ how would you split the above into partial fractions?

you just add and subtract two! so x+2-2/x-2
and this gives you 1+(2/x-2)
Original post by Tikara
exactly what I'm doing lmao - although I'm mainly just focusing on the big integration questions :P


ahaaa so it's not just me :smile:
Original post by Rayquaza
Sorry, no...Still lost. Like a cow on Astroturf.

Suppose that ex2e^{x^2} was a guess to the answer to the integral. To check if its right, you differentiate it to see if you get 6xex26xe^{x^2}. You get 2xex22xe^{x^2} instead. Its not exactly what you need, but its almost identical. How can you change the initial guess so when you do differentiate you get 6xex26xe^{x^2}.

Alternatively, a substitution of u=x2u=x^2. Its a longer way, but if you're more comfortable with it, then its the better way.

Original post by Myocardium
∫xx−2 dx\displaystyle \int \frac{x}{x-2} \ dx

QQ how would you split the above into partial fractions?


xx−2=x−2+2x−2=x−2x−2+2x−2=1+2x−2\dfrac{x}{x-2} = \dfrac{x-2+2}{x-2} = \dfrac{x-2}{x-2}+\dfrac{2}{x-2} = 1+\dfrac{2}{x-2}
Reply 5099
Avatar for ITomI
ITomI
Original post by Myocardium
∫xx−2 dx\displaystyle \int \frac{x}{x-2} \ dx

QQ how would you split the above into partial fractions?


long division, hope this helps,
kShVPQF.jpg50.5KB

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