## AravindG Group Title Find the points of maxima/minima of 2 years ago 2 years ago

1. hal_stirrup

whats you Q?

2. AravindG

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

3. AravindG

@amistre64 , @UnkleRhaukus , @Callisto , @Ishaan94

4. AravindG

@.Sam.

5. AravindG

@cwrw238

6. UnkleRhaukus

(t2−5t+4)=(t-1)(t-4)

7. hal_stirrup

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

8. AravindG

?

9. AravindG

help pls

10. hal_stirrup

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

11. AravindG

maxima and minima !!

12. cwrw238

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

13. Spacelimbus

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 (-:

14. 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

15. cwrw238

yea spacelimbus - im not sure either!

16. hal_stirrup

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

17. Spacelimbus

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

18. Spacelimbus

@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?

19. hal_stirrup

yes

20. Spacelimbus

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

21. Spacelimbus

is the derivative of the integrated function*

22. cwrw238

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

23. cwrw238

thats some integral!!!! lol

24. hal_stirrup

ok integral is the area under the curve.

25. cwrw238

yes

26. Spacelimbus

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*

27. 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

28. Spacelimbus

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.

29. hal_stirrup

typo error cwrw238 i think too.

30. cwrw238

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

31. cwrw238

typo?

32. Spacelimbus

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

33. 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...

34. hal_stirrup

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 .

35. hal_stirrup

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