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AravindG
 3 years ago
Find the points of maxima/minima of
AravindG
 3 years ago
Find the points of maxima/minima of

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

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 , @UnkleRhaukus , @Callisto , @Ishaan94

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0(t2−5t+4)=(t1)(t4)

hal_stirrup
 3 years ago
Best ResponseYou've already chosen the best response.0am not sure if you need to know the answer or how to go about answering the question.

hal_stirrup
 3 years ago
Best ResponseYou've already chosen the best response.0so i can help. i need to know what are you really looking for?

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1the expression in the integral = 0 for max/minm points ( i think) isn't that right guys?

Spacelimbus
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 (:

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1so (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

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1yea spacelimbus  im not sure either!

hal_stirrup
 3 years ago
Best ResponseYou've already chosen the best response.0integral (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

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

Spacelimbus
 3 years ago
Best 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?

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

Spacelimbus
 3 years ago
Best ResponseYou've already chosen the best response.0is the derivative of the integrated function*

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1makes sense to me though i'm learning calculus at present time so I'm not expert by any means

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1thats some integral!!!! lol

hal_stirrup
 3 years ago
Best ResponseYou've already chosen the best response.0ok integral is the area under the curve.

Spacelimbus
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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*

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1nor 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

Spacelimbus
 3 years ago
Best ResponseYou've already chosen the best response.0Seems 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.

hal_stirrup
 3 years ago
Best ResponseYou've already chosen the best response.0typo error cwrw238 i think too.

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1oh yes of course  i only differentiated the numerator!!! my mistake there.

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

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.1i 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...

hal_stirrup
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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 .

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