Well you know that by the properties of integrals, you can treat it as two integrals
I=I_1+I_2=\left(\int 1dt\right)+\left(-\int\cos{t}dt\right)
Do you know these integrals? What is the derivative of -\sin{t}?
There are two parts two a problem like this. First you'll want to find the direction of the line of intersection, which is nothing but the cross product of the normal vectors of the planes, i.e. \mathbf{n_1}\times\mathbf{n_2}=\langle 2,-1,-1\rangle \times \langle 1,2,3\rangle. Then all you need...
To solve differential equations in the form
y'+P(x)y=Q(x)
it is useful to use an integrating factor defined by
\mu=\exp{\left(\int P(x)dx\right)}
We multiply both sides of the equation by this,
\mu y'+\mu P(x)y=\mu Q(x)
and if you look closely the left hand side is the...
Well if you have that
\sin{(4t)}=2\sin{(2t)}\cos{(2t)}
then
\sin{(2t)}\cos{(2t)}=\frac{1}{2}\sin{(4t)}
So then if you want to do it from the integral you just need to integrate
\int_0^{\infty}\frac{1}{2}\sin{(4t)}e^{-st}dt
where the s you treat as a constant (you're...
You've pretty much finished it, all you need to recognize that
Y(s)=\frac{s-1}{(s-1)^2+1}
through completing the square in the denominator. Now can you get that to work with
f(t)=L^{-1}\left\{\frac{s-a}{(s-a)^2+b^2} \right\} = e^{at}\cos{(bt)}
You are right about the local max and min. For the second part--all an inflection point is, is where the second derivative (that which we use to determine concavity) is 0. So you would have
f''(x)=2(x-2)+4(x+1)=0
Can you get it from there?
It looks fine except I would check your order of limits--it should be entering at x=0 and leaving at x=y-2 and it's backwards for y too--maybe that was just a mistake when you wrote it.
Don't go back to sin and cos it's much easier to deal with the exponential terms. Sorry I didn't mention it but it looks like they are Gaussian, you want to get it in the form \int_{-\infty}^{\infty}e^{-(x-a)^2} where this is a Gaussian integral centered at a.
So you're dealing with right now...
Yea it does take awhile; it looks like the first 4 non-zero terms are (it takes 21 terms to get them)
1-4x^7+10x^{14}-20x^{21}
That looks like \sum a_n(-1)^nx^{7n}
where the [itex]a_n[/tex] is some sequence that makes the {1,4,10,20,...}; I can't think of what would work for that at...
Well this basically looks like an off-center ellipse as you can see from the x term, so you'll want to parametrize it using the trig identity \cos^2x+\sin^2x=1
What happens if you complete the square on the x term and make it as such?...
y=\sqrt{\frac{21}{2}}\sin{t}...
I think there is an error in your polynomial division if what you are meaning to say is that
\frac{2x^3}{x^3-1}=2+2\frac{1}{x^{-1}}=2+2x
The way that I would do it is to split up the integral like this..
\int\frac{2x^3}{x^3-1}dx=\int x \frac{2x^2}{x^3-1}dx=x\ln{(x^3-1)}-\int \ln{(x^3-1)}dx...
Think of the cos term and the fact that
\cos{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2}
Then using that you should be able to solve the integral.
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{e^{ix^2}+e^{-ix^2}}{2}e^{-ikx}dx
It looks like it is first-order homogeneous. Make the substitution z=x/t by dividing by t throughout your t/(2t+x), and then x=tz and x'=z+tz' by the product rule. Now it's in a form you want it for the relevant equations you listed. Just make the substitution the other way when you've solved for z.
Your matrix A reduces to the identity matrix in reduced row echelon form; so then the column space is made up of all the columns of the original matrix;
\text{Col}(A)=\left\{
\begin{pmatrix}
1 \\
-3
\end{pmatrix}
,
\begin{pmatrix}
2 \\
5
\end{pmatrix} \right\}
So does the vector they're...
That would be the parametrization of the line y=x, but your curve that the line integral is over is the ellipse
4x^2+25y^2=100 \Rightarrow \text{ } \frac{2^2}{10^2}x^2+\frac{5^2}{10^2}y^2=1
I wrote it in a more suggestive way; can you see why the parametrization should be the...