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anonymous
 4 years ago
integration question
anonymous
 4 years ago
integration question

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{3/2}^{1/2} dx/4x^{2}+12x+13\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3complete the square in the denominator and then trig substitute

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i tried that but.. got stuck at the +13

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.08 is not a square and if u factorise (x+2)^2 u dont get 4x^2....

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3factor that four out of the denom first\[\int_{\frac32}^{\frac12}\frac{dx}{4x^2+12x+13}=\frac14\int_{\frac32}^{\frac12}\frac{dx}{x^2+3x+\frac{13}4}\]\[=\frac14\int_{\frac32}^{\frac12}\frac{dx}{x^2+3x+\frac94+1}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3\[=\frac14\int_{\frac32}^{\frac12}{dx\over(x+\frac32)^2+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did u get 9/4+1? =/

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.313/4=(9/4)+1 completing the square on x^2+3x means adding a term of 9/4 you can save some trouble by noticing that 13/4=1+9/4 so we can make our constant term from the fraction we already have

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3trig sub:\[x+\frac32=\tan\theta\implies dx=\sec^2\theta d\theta\]change bounds...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338250956092:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to be honest i just want to know how my teacher got dx/4x^2+12x+13 to be dx/(2x+3)^2+2^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[4x ^{2}+12x+13 \to (2x+3)^{2}+2^2\] can u tell me how he got it =/..thx

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3complete the square\[4x^2+12x+13=4x^2+12x+9+4=(2x+3)^2+2^2\]knowing that splitting the terms up like this is not as obvious as the way in which I did it I think, so it requires a little more intuition I prefer to get the coefficient of the x term to be 1 to be safe, but notice that you can take my answer and your teachers result follows if you let the 4 back in the denom

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3he just noticed that splitting the 13=9+4 would work

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3above I derived it my way\[=\frac14\int_{\frac32}^{\frac12}\frac{dx}{x^2+3x+\frac94+1}\]let back in the 4 now...\[=\int_{\frac32}^{\frac12}\frac{dx}{4x^2+12x+9+4}\]from whence follows your teacher's way note both will give the same answer obviously

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright thx.. but the way u did is not in my course =/ i dont even know what changing bounds means..yet lol

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3the bounds in terms of x are 3/2 and 1/2 we make the substitution \[2x+3=2\tan\theta\]plug in the bounds of x and we can get them in terms of theta\[2(\frac32)+3=0=2\tan\theta\implies \theta=0\]\[2(\frac12)+3=2=2\tan\theta\implies\theta=\tan^{1}1=\frac{\sqrt2}2\]so the new bounds are\[\int_{0}^{\frac{\sqrt2}2}\text{(whatever)}dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0:o..never did that.. before :P..kinda confusing.. we jsut call what u call bounds..limits

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3whatever floats your boat I just don't like getting that term confused with limits like\[\lim_{x\to a}f(x)\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3call 'em Freddies for all I care though, makes no mathematical difference whatever is easiest for you to keep straight

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol.. ya i guess, thx again see u around

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3welcome, good luck :)
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