Homework Statement
Factor.
2x^{2}+5x-12
I just took a semester off from school and I feel dumb. My recommendation to anyone reading is don't do that.
Anyways, back to gr. 10 math :cry:
Normally if I was going to factor this I would try to eliminate the coefficient of x^{2} but it...
Ok, should it go something like this?:
\int_{1}^2cos(px)dx
Let u = px
Therefore, du = pdx
And, dx = \frac{du}{p}
So,
\int_{1}^2cos(px)dx = \int_{1}^2cos(u)\frac{du}{p}
= \frac{sin(u)}{p} + c
Is that correct?
Exactly. So then this is my logic:
Since sin^2(u) = [sin(u)]^2
Let u = 3x
And let v = sin(u)
i.e.
\frac{d}{dx}v^2 = 2v = 2(sin(u)) * \frac{d}{dx}sin(u) = 2(sin(3x)) * cos(3x) * \frac{d}{dx}3x
= 2(sin(3x)) * cos(3x) * 3
= 6(sin(3x)cos(3x))
And since...
Homework Statement
Solve the following Integral:
\int_{1}^2cos(px)dx
where p is a constantHomework Equations
The Attempt at a Solution
I'm totally lost here...
Oh man, I really over complicated things. Thanks :D
EDIT: Wait though... shouldn't sin switch to cos at some point?
Doesn't the chain rule mean it should go something like this:
= 2sin(3x)*cos(3x)*3
= 6* sin(3x)cos(3x)
= 3( 2sin(3x)cos(3x) )
= 3( sin(2 *3x) )
= 3sin(6x)
Hi, I've got to solve for: \int\stackrel{3}{1}(4x2+2)dx
This is what I've done:
= [ (4/3)x3 + 2x ]\stackrel{3}{1}
= [ (4/3)(3)3 + 2(3) ] - [ (4/3)(1)3 + 2(1) ]
then I solved...
= 44/3
Is that correct at all?
I'm not sure how to differentiate sin^2[3x]. Although, I think it's just d/dx( (sin[3x])(sin[3x]) ). So, just chain and product rules should do it. Is that right?
EDIT: I've followed through with the above method, and I got 3*sin(6x). Is that correct?
Thanks everyone. I actually do understand how to get this particular example's anitderivative. I wanted to know if there was a more general way of achieving this. I can only imagine once f(x) gets a little more complicated, finding the antiderivative could get quite painful (although finding...
This is out of my textbook:
EXAMPLE 6:
Find the area under the parabola y=x2 from 0 to 1.
SOLUTION:
An antiderivative of f(x) = x2 is F(x) = 1/3x3. The required area is found using Part 2 of the Fundamental Theorem...
My question is: how was the antiderivative obtained?
It's nothing to do with me; even my professors and TAs have said it's a problem with the high school curriculum. They no longer sufficiently prepare students for university... one of the most lacking subjects apparently is math.
With that said, thank you all for your advice. I am enrolled...