Calculus AP AB help, please!!!

I need help solving the following
1. integral((16x dx)/sqrt(8x^2+1))

I know the derivatives of inverse trigonometric functions but this doesn't look like anyof them...I mean none of them have a square root of two numbers added together, on the bottom.

 

Since 16x is in the numerator, I think all you have to do is raise the quantity in the denominator like a normal function. I'll work on it and let you know if I get an answer.

Edit - Yeah, the answer is just sqrt(8x^2 + 1)/2 + C. I think. You may want to check that.

Amazing.

 

2*sqrt(8x^2 + 1) + C

close amazing.

 

That's what I meant.

Amazing.

 

Uhm....one of you guys want to tell me how you got that?

 

It's hard to explain on a message board. And since I can't remember how to get the times 2 part, I will let someone else explain. But it all has to do with the 16x in the numerator, and the fact that the derviative of 8x^2 is 16x, so all you have to do if integrate the denominator as you would a regular function.

Amazing.

 

Here ya go:

First look at the function. Try to break it up into parts. Here you have 16x and sqrt(8x^2 + 1). Next try to see if one of them relates to the other. You can see that 16x is the derivative of 8x^2 + 1. Once you know that one "part" is the derivative of another "part" you can figure that the chain rule was used. Now just take the... argh here I'll explain it mathematically.

you have

dy = ______16x______dx
``````sqrt(8x^2+1)

you notice that the derivative of 8x^2+1 is 16x. So you rename 8x^2+1 as u. therefore

u = 8x^2+1
du = 16x*dx

now replace the original equation by du and u's.

dy = ______1______du
``````sqrt ( u )

1/sqrt(u) is pretty ease to differentiate, it is simply 2*sqrt(u).

So now you have

y = 2*sqrt(u) = 2*sqrt(8x^2+1)

I hope that helped.

 

I got it now..thanks guys.

 

before or after I did all that typing?

 

Yeah I figured it out...lol I feel so stupid now. The whole rest of the section was integrals that get you inverse trigonometric functions though so I didn't even think of a normal integral.

 

Before Sorry. And yeah...I've got more questions.(Note to self: don't do calc homework at 12:30 am when it's too late to call classmates for help)
Oh and if you do help again I don't need all the text inbetween the steps. If you just post the different steps then I'll figure out what you're doing. Anyway....

"Evaluate integrals in Exercises 23-28 by completeing the square and using substitution to match each to a standard form."

25. integral ((8dx)/(x^2-2x+2))

....I hate completing the square

29. integral ((cscx - cotx)^2 dx)

 

25. is 8arctan(x-1)+C, I'll type up the steps in a second.

Steps
I'm going to omit the integral signs cause they're hard to type.

Complete the square so you end up with: 8dx/(x^2-2x+1)+2-1.

Simplify to: 8dx/(x-1)^2+1.

Pull out the 8 so you have 8 integral dx/(x-1)^2+1.

Let U= x-1
dU=dx

so: 8 integral du/u^2+1, which equals 8arctan(U)+C, plug in x-1 for U, to get 8arctan(x-1)+C.

 

Answer X+1/sinX=X+cscX

1. carry out square
(cotX)^2-2cscX*cotX+(cscX)^2

2. Simpfy terms

-2cscX*cotX=-cosX/(sinX)^2 (cscX=1/sinX, cotX=cosX/sinX)

cotX^2(->=(cotX)^2)=1-(1/sinX^2)=1-cscX^2 (cosX^2=1-sinX^2)

3. Chain Rule

You now have Intergal((cscX^2+1-cscX^2)dx)+Intergal(-cosXdX/sinX^2) cscX^2 cancel so you have intergal of 1=X and cosXdX=dU and U=sinX so now the second one is Inttergal(-dU/U^2)=1/U->1/(sinX)

4. Answer X+1/sinX=X+cscX

 

Well I'm back

2. Let f be the function given by f(x) = 2xe^(2x)

a) Find the lim of f(x) as x--> (negative infinity) and lim of f(x) as x--> (infinity)

b) Find the absolute minimum value of f Justify that your answer is an absolute minimum.

c)What is the range of f?

d) Consider the family of functions defined by y= bxe^(bx), where b is a nonzero constant. Show that the absolute minimum value of bxe^(bx) is the same for all nonzero values of b.

Thanks a lot!
------------------------
Edit:Note that <= means "less than or equal to" The graph of the velocity v(t) in ft/sec of a car traveling on a straight road for 0<=t<=50 is shown above. A table of values for v(t) at 5 second intervals of time t is shown to the right of the graph.

Here is the table:
t(seconds) v(t)
0 0
5 12
10 20
15 30
20 55
25 70
30 78
35 81
40 75
45 60
50 72

b) find the average acceleration of the car, in ft/sec^2 over the interval 0<=t<=50

c) Find one approximation for the acceleration of the car in ft/sec^2 at t=40 Show the computations you used to arrive at your answer

 

anyone??

 

2. f(x) = 2xe^2x
a) as x approcahes negative infinity f(x) approaches zero
as x approcahes infinity f(x) approches infinity
b)stat pt f'(X)=0
f'(x)= 4xe^2x + 2e^2x
therefore 4x+2=0 (for stat pt)
x=-1/2
Test Stat pt
x=-1, f'(x)= -0.270
x=-1/2 f'(x)=0
x=0 f'(x)=2
therefore since curve cts and gradient changes from (-,0,+) therefore relative min. and since as x approaches minus infinity f(x) approcahes 0 and as x approches infinity f(x) approcahes infinity, therefore also absolute min.
c)f(x) at x=-1/2 =<f(x)=<infinity

d)Let f(x)=bxe^bx
f'(x)= e^bx(b+xb^2)
stat pt is f'(x)=0
x=-1/b
sub x=-1/b into f(x) and you get f(x)=-1/e (indepenfdant of b) therefore always same min value


Next qu
a) a=(v-u)/t
a= 72/50 m/s^2

b) instantaneous acceleration (graph the points and draw a line of best fit and then find gradient at t=40)

 

For part C of the second question, I think you can use the average acceleration between t=35 and t=45 (take the secant line and numerically calculate the slope).