Thread: integrals and implicit - and fast

1. integrals and implicit - and fast

ok help: we're given the following:
$\int_{0}^{5}{f(x)dx = 8}$
f(x) is even function
$\int_{0}^{5}{g(x)dx = 4}$
g(x) is odd function

find the following (if possible):
$\int_{0}^{5}\frac{f(x)}{g(x)}dx$

next:
$f(x) = \int_{1}^{2x}\frac{dt}{(t^2+1)^\frac{1}{3}}$
find f'(1)
(I feel as though $\frac{1}{2^\frac{1}{3}}$ would be too easy)

next:
let ${x^2 - 4xy + y^2 = 3}$, find expression for $\frac{dy}{dx}$
when I do it I get $\frac{2y-x}{y-2x}$
then it goes on asking "are there any points on curve where tangent is horizontal or vertical?"

2. Originally Posted by cassiopeia1289
ok help: we're given the following:
integral from 0 to 5 of f(x) = 8
f(x) is even function
integral from 0 to 5 of g(x) = 4
g(x) is odd function

find the following (if possible):
integral from 0 to 5 of [f(x)/g(x)]dx

next:
f(x) = integral from [1 to 2x] of [dt/(t^2+1)^(1/3)]
find f'(1)
(I feel as though 1/(2^(1/3)) would be too easy)

next:
let x^2 - 4xy + y^2 = 3, find expression for dy/dx
when I do it I get (2y-x)/(y-2x) where (y-x) cancel out and I just get 1 - which doesnt even make sense

then it goes on asking "are there any points on curve where tangent is horizontal or vertical?" well, obviously its a constant then ...
I have to have done something wrong

(AS WE SPEAK - WORKING ON TYPING THESE IN USING ACTUAL MATH FORM - JUST BE PATIENT, LOL)
The second one you have $\int_a^{2x}f(t)dt$ the derivative would be $2f(2x)$

3. Hello,

For the second one, let G(t) be an antiderivative for $g(t)=\frac{1}{(t^2+1)^{1/3}}$

According to the definition of an integral, $f(x)=G(2x)-G(1)$

Hence, $f'(x)=2g(2x)$

So $f'(1)=2g(2)=2 \frac{1}{5^{1/3}}$

4. let x^2 - 4xy + y^2 = 3, find expression for dy/dx
when I do it I get (2y-x)/(y-2x) where (y-x) cancel out and I just get 1 - which doesnt even make sense
Well...

$\frac{2y-x}{y-2x}=\frac 12 \times \frac{y-\frac x2}{y-2x}$

I don't find that it cancels out

5. ok - yeah it doesnt cancel - but I finally figured out that horiz tangent at $x^2$ = -4
but that doesn't work, so there is no horizontal tangent?

6. I had the expression for $f'(x)$ so I just replaced x by 1..

7. (unrelated - but how do you write definite integrals with the numbers there?)

8. \int_{bbbbbb}^{aaaaaa}

9. can you help with the first and last as well?

10. Originally Posted by cassiopeia1289
ok help: we're given the following:
$\int_{0}^{5}{f(x)dx = 8}$
f(x) is even function
$\int_{0}^{5}{g(x)dx = 4}$
g(x) is odd function

find the following (if possible):
$\int_{0}^{5}\frac{f(x)}{g(x)}dx$

next:
$f(x) = \int_{1}^{2x}\frac{dt}{(t^2+1)^\frac{1}{3}}$
find f'(1)
(I feel as though $\frac{1}{2^\frac{1}{3}}$ would be too easy)

next:
let ${x^2 - 4xy + y^2 = 3}$, find expression for $\frac{dy}{dx}$
when I do it I get $\frac{2y-x}{y-2x}$
then it goes on asking "are there any points on curve where tangent is horizontal or vertical?"
$2x-4[xy'+y]+2yy'=0$

solve for y'

11. Tangent is horizontal when y'=0 and vertical when y' tends to infinity.

So isolate y' and find for which values of x the conditions are verified.

For the first one, I think it's not possible with such elements

12. Originally Posted by Moo
Tangent is horizontal when y'=0 and vertical when y' tends to infinity.

So isolate y' and find for which values of x the conditions are verified.

For the first one, I think it's not possible with such elements
I agree with Moo...the first one is impossible unless it wanted $\frac{\int_0^{5}f(x)dx}{\int_0^{5}g(x)dx}$

But if it was I am sure you could figure it out

13. ok, good to know about the first one

the last one - I solved and found that I would have to take the square root of a negative, since no imaginary numbers would work for vertical or horizontal tangents, I'm kinda stuck - so are there just none?

14. Originally Posted by cassiopeia1289
^ yeah, and when I did I got what I put below in the original post
what I don't know is how to find the horizontal and vertical tangent point/line things
Horizontal tangent lines occur when the first derivative is 0

15. ^ yes, I realize this - take the derivative yourself and you'll see my problem

$\frac{dy}{dx} = \frac{2y-x}{y-2x}$
now you get $0 = \frac{2y-x}{y-2x}$
I solved for y and got $y = \frac{x}{2}$
I plugged that into the original problem so as to find the x solution and therefore find the horizontal line
so: $x^2 - 4x(\frac{x}{2}) + (\frac{x}{2})^2 = 3$
solve and you get $x^2 = -4$
this only works if you use imaginary numbers, so is there just not a horizontal tangent line?

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