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Mathematics 8 Online
OpenStudy (anonymous):

Taking the derivative of an integral...

OpenStudy (anonymous):

How do I take the derivative of the function \[g(x)=\int\limits_{x}^{x^2}(t^2-5)/(t^2+2)dt\]

OpenStudy (amistre64):

the derivative and an integral are inverse operations: thed cancel each other out and youd end up with: g'(x) = (t^2-5)/(t^2+2) ..... as far as I recollect

OpenStudy (anonymous):

How exactly does it cancel out?

OpenStudy (amistre64):

hmmm..... you might have a point, I failed to recognize the variables....

OpenStudy (anonymous):

So.. how would you do it?

OpenStudy (amistre64):

take the integral, and substitute for t with x...... then derive it back down..... but the actuall process eludes me :)

OpenStudy (amistre64):

can we get the integrand into a sum of terms?

OpenStudy (amistre64):

dunno if a trig substitution would be useful: t = sqrt(2) tan(a)

OpenStudy (anonymous):

Idk.. I'm just totally lost on this question. So I substitute the t's for x's?

OpenStudy (anonymous):

Or do I substitute the t^2 in?

OpenStudy (amistre64):

my limited knowledge on this would suggest that I find g(x) from the integrand first.... an exact HOW, is beyind me at the moment :)

OpenStudy (anonymous):

Lol ok. Thank you for your help though :)

OpenStudy (amistre64):

if I did it right in my thoguths here: the new integrand becomes: tan^2(a) -5/2 which looks easier to work with...

OpenStudy (anonymous):

nicole did you finish most of the packet? i didn't...i do have a much geater understanding about integrals though :)

OpenStudy (anonymous):

No.. I think I'm gonna stay after school.

OpenStudy (anonymous):

mother just told me she is picking me up. i have NO idea why. i want to stay too..

OpenStudy (amistre64):

i substituted: t = sqrt(2) tan(a) ; dt = sqrt(2) sec^2 2tan^2(a) -5 sqrt(2)sec^2(a) da ------------- ------------- 2(tan^2(a) + 1) 1 2tan^2(a) -5 sqrt(2)sec^2(a) da ------------- ------------- 2sec^2(a) 1 2tan^2(a) -5 sqrt(2) da ------------- ------- 2 1 2tan^2(a) -5 sqrt(2) da ---------- - --- ------- 2 2 1 sqrt(2) tan^2(a) - 5sqrt(2)/2 da is what I get actually, but it might be easier to integrate

OpenStudy (anonymous):

Just stay.. find a ride home.

OpenStudy (anonymous):

Wow.. That's a whole lot..

OpenStudy (amistre64):

or since the sqrt(2) is a constant, we can yank it out to the other side: sqrt(2) [S] tan^2(a) - 5/2 da

OpenStudy (amistre64):

myins here....now ill now what Im doing wrong ;)

myininaya (myininaya):

Let f(t)=(t^2-5)/(t^2+2) Then int(f(t), t=x..x^2)=F(x^2)-F(x) To take derivative of this. We have to use chain rule. Therefore g'(x)=2xf(x^2)-f(x)

myininaya (myininaya):

g'(x)=2x[(x^4-5)/(x^4+2)]-(x^2-5)/(x^2+2)

myininaya (myininaya):

any questions?

myininaya (myininaya):

we dont need to integrate to find the derivative of the integral

OpenStudy (amistre64):

......thats what I thought :) but those x..x^2 things threw me for a loop

myininaya (myininaya):

lol where did you get all of those fans? I'm so jealous

OpenStudy (amistre64):

they were having a sale at pier1 ;)

myininaya (myininaya):

damn i missed it :(

myininaya (myininaya):

i think he lefted :(

OpenStudy (amistre64):

maybe, but the math remained lol

myininaya (myininaya):

true im going to go play call of duty now peace

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