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AravindGBest ResponseYou've already chosen the best response.0
\[\int\limits^{x^2}_0 [(t^25t+4)/(2+e^t)]dt\]
 one year ago

AravindGBest ResponseYou've already chosen the best response.0
@amistre64 , @UnkleRhaukus , @Callisto , @Ishaan94
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
(t2−5t+4)=(t1)(t4)
 one year ago

hal_stirrupBest ResponseYou've already chosen the best response.0
am not sure if you need to know the answer or how to go about answering the question.
 one year ago

hal_stirrupBest ResponseYou've already chosen the best response.0
so i can help. i need to know what are you really looking for?
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
the expression in the integral = 0 for max/minm points ( i think) isn't that right guys?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
I would have guessed so @cwrw238, I am not sure if the integrated function and the to be integrated function share similarities in minima and maxima though. but for linear functions that seems to be the case (:
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
so (t  4)(t  1) = 0 t = 4 or 1 so you have turning points at values t = 4 and t = 1 second derivative is 2t  5 when t = 4 this is positive so t=4 is a minimum when t = 1 this is negative so t = 1 is a maximum
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
yea spacelimbus  im not sure either!
 one year ago

hal_stirrupBest ResponseYou've already chosen the best response.0
integral (45 t+t^2)/(2+e^t) dt = (5/2t) Li_2(e^t/2)+Li_3(e^t/2)+t^3/6(5 t^2)/41/2 t^2 log(1/2 (e^t+2))+2 t+5/2 t log(1/2 (e^t+2))2 log(e^t+2)+constant
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
THe polylogarithmic function behaves weirdly for min/max though @hal_stirrup , I wouldn't know how to solve that expression for zeros.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
@cwrw238, lets think about this together, an integral is the antiderivative of a function right? And to find minima and maxima you need to set the derivative to zero, correct?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
therefore, what we have next to the integral expression is the derivative of the function, I would say your answer is pretty much correct.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
is the derivative of the integrated function*
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
makes sense to me though i'm learning calculus at present time so I'm not expert by any means
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
thats some integral!!!! lol
 one year ago

hal_stirrupBest ResponseYou've already chosen the best response.0
ok integral is the area under the curve.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Oh okay @cwrw238, I believe it's correct, it's how I would read it anyway, it's basically an easier way (that looks more complicated at first glance though) If you integrate an arbitary function you get F(x), if you derive that again you get f(x), the derivative of a function is what decides if there are minima and/or maxima points present. So in our case, just the f(x) =0, because that is the derivative of the obscure function I wouldn't know how to integrate without series and transformations *grins*
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
nor me!!  yes i think these values of t correspond to max ./ mini on the graph . just noticed the limits on integral  i expect that t^2 not x^2
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
Seems to be a time > position transformation, but I cannot tell, however, the points we have found are extremal points, but to decide wether they are min/max or not we need the second derivative of the function f(x) which is a bit more complicated since it requires the use of the quotient rule.
 one year ago

hal_stirrupBest ResponseYou've already chosen the best response.0
typo error cwrw238 i think too.
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
oh yes of course  i only differentiated the numerator!!! my mistake there.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
not a big thing, just some additional work (: but @AravindG can verify these points for himself.
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
i just entered the integral on wolfram alpha http://www.wolframalpha.com/input/?i=integrate+++%28t%5E2−5t%2B4%29%2F%282%2Be%5Et%29+  theres no min/max on the plot at the points t = 1 and 4 so it looks like i'm wrong...
 one year ago

hal_stirrupBest ResponseYou've already chosen the best response.0
ok i recommend 1: to go back the question and se if there is a typo error. 2: go back to basic integration methods in definite integral . and you will be able to solve it .
 one year ago

hal_stirrupBest ResponseYou've already chosen the best response.0
example :\[\int\limits_{}^{} x^4\] its primative funaction is : \[x ^{5}/5\]
 one year ago
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