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zeppBest ResponseYou've already chosen the best response.0
1/cos^2(x) is sec^2(x) and the integral of that is tan(x)
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{2}x \sqrt{92x^2}dx\]
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
\[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\]
 11 months ago

primeralphBest ResponseYou've already chosen the best response.1
no, just substitute, then everything falls out
 11 months ago

Best_MathematicianBest ResponseYou've already chosen the best response.0
use calculator so easy
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
I need to do it without @best_mathematician
 11 months ago

Best_MathematicianBest ResponseYou've already chosen the best response.0
ok here...
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
integral and derivative cancel each other out
 11 months ago

Best_MathematicianBest ResponseYou've already chosen the best response.0
now u know wht to do right @Jgeurts
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
@modphysnoob show me please
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
\[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\] after our supposed integration F(Sin(x))  F ( x^2) take dervaitve d/dx of F( Sin(x)) = cos(x) f(sin(x) d/dx of F(x^2) = f (x^2) 2x go back and plug into your funciton
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
f(t)= Sqrt[1+t^2] f(sin(x)= sqrt[1+ sin^2(x)] f(sin(x)*cos(x)= sqrt[1+ sin^2(x)]* cos(x)
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
now the bottom part f(x^2)2x=sqrt[1+x^2]2x
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
sqrt[1+ sin^2(x)]* cos(x)sqrt[1+x^2]2x
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
Im formulating my question, haha
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
so the d/dx means derive the interval?
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
yes, since we are taking a function integrate it and derive , it undoes the integration
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
haha great! i get it, thank you, that one was super hard!
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
I can show you an easy example if you like?
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
@modphysnoob sure that would be great, im learning for my final :)
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
let's to a problem like yours but with easier integrand dw:1367723123193:dw
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
so we would derive sinx to cosx and x2 to 2x
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
then plug in for t and subtract them from each other right?
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
so in this case f(t)= t when we integrate we will get a function which we will call F(t) F(t) is evaluated at F(sin(x))  F( x^2) then taken d/dx of d/dx( F(sin(x))  F( x^2) ) using chain rule d/dx ( F(sin(x)) = cos(x) f( sin(x)
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
which is cos(x) sin(x)
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
let's do botttom part d/dx( F (x^2) = 2x f( x^2) 2x x^2= 2x^3
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
so cos(x) sin(x)2x^3
 11 months ago

modphysnoobBest ResponseYou've already chosen the best response.2
this is easy integrand so we could actually integrate it and then differntiate it to confirm out result. Integrate t t^2/2 plug in limits (sin^2 x)/2  x^4/2 take derivative d/dx cos(x) sin(x) 2x^3
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.0
Yes thats great! thank you so much! @modphysnoob
 11 months ago
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