Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 Re: Help Me ! » evaluate definite integral » 2008-06-29 03:43:57
Thanks again luca-deltodesco:)
#2 Help Me ! » Initial-value » 2008-06-29 03:40:58
- baggins
- Replies: 1
Hi everyone,
Just to be sure I did that problem with no major mistakes?
Here is the question and my answer:
Solve the initial-value problem
dy/dx = e^2x + 1/e^2x + 2x +2 (x>0), y= 3 when x= 0.
Y = ∫ e^2x + 1 dx General solution
e^2x + 2x +2
∫ f(x) dx = In(f(x)) + c (integration formula)
f(x)
The integrand is e^2x + 1/e^2x + 2x +2 .
Here the numerator e^2x + 1 is, except for a constant multiple, the derivative of the denominator e^2x + 2x +2 .
So we can apply equation with f(x) = e^2x + 2x +2 .
Since f(x) = 2e^2x+1x, we write the numerator as e^2x + 1= ½(2e^2x+1x) before applying the formula.
Thus we have
∫ e^2x + 1 dx = 1/2 ∫ 2e^2x+1x dx
e^2x + 2x +2 e^2x + 2x +2
= ½ In(e^2x + 2x + 2) + c
Where c is an arbitrary constant.
Using the initial condition, y= 3 when x= 0, we obtain
3= ½ In (e^0 + 0 +2) + c = c
c= 5/7
y= ½ In (e^2x + 2x +2) + 5/7
I hope it's not too bad:/
#3 Re: Help Me ! » evaluate definite integral » 2008-06-28 21:39:24
Hi luca-deltodesco,
Thanks for your help, how did you write it like this, is there a link on the website on how to use symbols?:)
#4 Help Me ! » evaluate definite integral » 2008-06-27 22:57:38
- baggins
- Replies: 4
Hi everyone,
I have to evaluate a definite integral but I'm stuck to finish it
Here is the question and my answer:
(b) Evaluate ∫3 on the top and 1 on the bottom( i don't how to write on the keyboard) follow by x(7x²+5)dx
= [7x³+5] 3 at the top and 1 at the bottom= (7x³+5)³ - (7x3+5)^1
I hope someone understand this, I don't know how to write the limits of integration with my keyboard.
:
Thanks:)
#5 Re: Help Me ! » derivative of the function using the Composite Rule » 2008-06-25 06:48:25
Thanks again mathsyperson, great help, I need to practise more of them.
#6 Re: Help Me ! » derivative of the function using the Composite Rule » 2008-06-25 03:11:51
Becarefule I type it wrong, the function 'k(x)' is k(x) = e^cos(6x)
#7 Help Me ! » derivative of the function using the Composite Rule » 2008-06-25 03:09:44
- baggins
- Replies: 3
Hi everyone,
I'm stuck, I'm sure I calculate the function completely wrong. I think cos is right but I don't know how to work the '6x'.
Here is the question and my answer:
(i) Write down the derivative of the function
f(x)= cos(6x).
Derivative:
f (x)= -sin(6x)
(ii) Hence, by using the Composite Rule, differentiate the function
k(x) = ecos(6x)
Composite rule:
k (x) = g (f(x)) f (x)
= g (u) f (x) where u = f(x) = cos(6x)
= (eu) (-sin^cos(6x))
= e(cos^(6x)) (-sin^cos(6x))
= -e sin(^cos(6x))
If someone understand what I did, please help me:/
#8 Re: Help Me ! » derivative of the polynomial function » 2008-06-24 22:43:58
Thanks luca-deltodesco for your help:):D
#9 Help Me ! » derivative of the polynomial function » 2008-06-24 22:12:12
- baggins
- Replies: 2
Hi Everyone,
Always making small mistakes in my calculation but unfortunately that make me loose points:(
I would like to know if I didn't make any mistake on that question:
This question concerns the function
f(x)= x^3 +3x^2 − 9x+15.
Find the stationary points of this function:
The derivative of the polynomial function
f(x) = x^3 + 3x^2 9x + 15
f (x) = (3x^2) + 3(2x) 9(1)
= 3x^2 + 6x 9
= 3(x^2 + 2x 3)
The equation f (x) = 0 is equivalent to x^2+2x-3 = 0;
That is, (x+3)(x-1) = 0
Which give the stationary points:
X= -3 and x= 1
Thanks for your help:)
Pages: 1