Question

Definite Integral question in comment

Answer 1

Given \[f(x) = \int\limits_{0}^{x} \frac{ t^2-49 }{1+\cos^2(x) }\]
At what value of x does the local max of f(x) occur?

Answer 2

forgot the dt at the end.

Answer 3

Is that supposed to \(\cos^2t\) in the denominator?

Answer 4

yes sorry

Answer 5

Recall the fundamental theorem of calc:
\[\frac{d}{dx}\int_c^{f(x)}g(t)\,\mathrm{d}t=g(f(x))\times \frac{d}{dx}[f(x)]\]

Answer 6

i suppose you start with applying f'(x)= integrant in terms of x

Answer 7

yeah the first part of the FTC

Answer 8

i basically told what you need "f'(x)= integrant in terms of x"

Answer 9

integrant is what's inside the intergral

Answer 10

So I would use the first part of FTC to solve this? Would I be able to break down the quotient further to apply the integral?

Answer 11

why you want to break it down?
you need to find critical values so set f'=0
if you do denominator can never be zero, so you are only concerned with top expression

Answer 12

do you know what to do?

Answer 13

I think so, but thanks for the help!

Answer 14

ok, no problem

Answer 15

recall max occurs if f' changes from positive to negative

Answer 16

Actually, how do you evaluate \[\int\limits_{}^{} \frac{ t^2-49 }{ 1+\cos^2t } dt\]? I do not understand the notation of the given FTC.

Answer 17

\[f'(x)=\frac{x^2-49}{1+\cos^2x}\]
we set \[f'(x)=0 ~~\Longrightarrow x^2-49=0\]

Answer 18

you don't need to evaluate the integral for now

Answer 19

so critical points occur at x=7 and x=-7 correct?

Answer 20

Wait, so the derivative of f(x) is the integrant? Yes, I worked it out and the max occurs at -7.

Answer 21

yes in terms of x. your integral has a different parameter if you realize
fundamental theorem of calculus part I
\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
Recall the fundamental theorem of calc:
\[\frac{d}{dx}\int_c^{f(x)}g(t)\,\mathrm{d}t=g(f(x))\times \frac{d}{dx}[f(x)]\]
\(\color{blue}{\text{End of Quote}}\)

Answer 22

in your case the function f(x) is not composite of two or more functions
so you don't need chain rule

Answer 23

Ohhh, I understand now! Thanks for the help again!

Answer 24

so you said you found max, how did you do it?

Answer 25

I plugged in the values, x = 7 and x = -7 into the derivative.

Answer 26

in the derivative? but that will give you f'=0 ?

Answer 27

you need to find the sign of f'

Answer 28

look at the sign of f' before -7 and after -7 and same for 7

Answer 29

denominator is already positive since 1+cosx>0
so forgot that number line i did

Answer 30

So,
(-inf, -7) : decreasing
(-7, 0) : decreasing
(0, 7) : decreasing
(7, inf) : increasing