- AravindG

Find the points of maxima/minima of

- chestercat

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- anonymous

whats you Q?

- AravindG

\[\int\limits^{x^2}_0 [(t^2-5t+4)/(2+e^t)]dt\]

- AravindG

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## More answers

- AravindG

@.Sam.

- AravindG

- UnkleRhaukus

(t2−5t+4)=(t-1)(t-4)

- anonymous

am not sure if you need to know the answer or how to go about answering the question.

- AravindG

?

- AravindG

help pls

- anonymous

so i can help. i need to know what are you really looking for?

- AravindG

maxima and minima !!

- cwrw238

the expression in the integral = 0 for max/minm points ( i think)
isn't that right guys?

- anonymous

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 (-:

- cwrw238

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

- cwrw238

yea spacelimbus - im not sure either!

- anonymous

integral (4-5 t+t^2)/(2+e^t) dt = (5/2-t) Li_2(-e^t/2)+Li_3(-e^t/2)+t^3/6-(5 t^2)/4-1/2 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

- anonymous

THe polylogarithmic function behaves weirdly for min/max though @hal_stirrup , I wouldn't know how to solve that expression for zeros.

- anonymous

@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?

- anonymous

yes

- anonymous

therefore, what we have next to the integral expression is the derivative of the function, I would say your answer is pretty much correct.

- anonymous

is the derivative of the integrated function*

- cwrw238

makes sense to me -though i'm learning calculus at present time so I'm not expert by any means

- cwrw238

thats some integral!!!! lol

- anonymous

ok integral is the area under the curve.

- cwrw238

yes

- anonymous

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*

- cwrw238

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

- anonymous

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.

- anonymous

typo error cwrw238 i think too.

- cwrw238

oh yes of course - i only differentiated the numerator!!! my mistake there.

- cwrw238

typo?

- anonymous

not a big thing, just some additional work (-: but @AravindG can verify these points for himself.

- cwrw238

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...

- anonymous

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 .

- anonymous

example :\[\int\limits_{}^{} x^4\] its primative funaction is : \[x ^{5}/5\]

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