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abz_tech
integration question
\[\int\limits_{3/-2}^{1/-2} dx/4x^{2}+12x+13\]
complete the square in the denominator and then trig substitute
i tried that but.. got stuck at the +13
8 is not a square and if u factorise (x+2)^2 u dont get 4x^2....
factor that four out of the denom first\[\int_{-\frac32}^{-\frac12}\frac{dx}{4x^2+12x+13}=\frac14\int_{-\frac32}^{-\frac12}\frac{dx}{x^2+3x+\frac{13}4}\]\[=\frac14\int_{-\frac32}^{-\frac12}\frac{dx}{x^2+3x+\frac94+1}\]
\[=\frac14\int_{-\frac32}^{-\frac12}{dx\over(x+\frac32)^2+1}\]
how did u get 9/4+1? =/
13/4=(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
trig sub:\[x+\frac32=\tan\theta\implies dx=\sec^2\theta d\theta\]change bounds...
to be honest i just want to know how my teacher got dx/4x^2+12x+13 to be dx/(2x+3)^2+2^2
\[4x ^{2}+12x+13 \to (2x+3)^{2}+2^2\] can u tell me how he got it =/..thx
complete 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
he just noticed that splitting the 13=9+4 would work
above 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
alright thx.. but the way u did is not in my course =/ i dont even know what changing bounds means..yet lol
the 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\[\int_{0}^{\frac{\sqrt2}2}\text{(whatever)}dx\]
:o..never did that.. before :P..kinda confusing.. we jsut call what u call bounds..limits
whatever floats your boat I just don't like getting that term confused with limits like\[\lim_{x\to a}f(x)\]
call 'em Freddies for all I care though, makes no mathematical difference whatever is easiest for you to keep straight
lol.. ya i guess, thx again see u around
welcome, good luck :)