Question

Find the points of maxima/minima of

Answer 1

whats you Q?

Answer 2

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

Answer 3

@amistre64 , @UnkleRhaukus , @Callisto , @Ishaan94

Answer 4

@.Sam.

Answer 5

@cwrw238

Answer 6

(t2−5t+4)=(t-1)(t-4)

Answer 7

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

Answer 8

?

Answer 9

help pls

Answer 10

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

Answer 11

maxima and minima !!

Answer 12

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

Answer 13

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 (-:

Answer 14

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

Answer 15

yea spacelimbus - im not sure either!

Answer 16

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

Answer 17

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

Answer 18

@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?

Answer 19

yes

Answer 20

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

Answer 21

is the derivative of the integrated function*

Answer 22

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

Answer 23

thats some integral!!!! lol

Answer 24

ok integral is the area under the curve.

Answer 25

yes

Answer 26

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*

Answer 27

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

Answer 28

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.

Answer 29

typo error cwrw238 i think too.

Answer 30

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

Answer 31

typo?

Answer 32

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

Answer 33

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...

Answer 34

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 .

Answer 35

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