Expert: Paul Klarreich - 4/5/2006

Question
Also:

Evaluate using integral table, indicate formula number and simplify as much as possible.


S=integral sign
1.
a) S square root sign over x^2(x squared) + 9dx


b) S ln 4-over-5 * xdx


c) S 2-over-5x(7x+2) * dx


d) S x^3e^-2x * dx



2. Determine whether the improper integral is convergent or divergent, and calculate its value if it is convergent (a & c)

a) S(infinity sign at the top; 0 at the bottom) 4e^-4x*dx

b) S(infinity sign at the top; 5 at the bottom) x^4*dx

c) S(infinity sign at the top; 1 at the bottom) 3x^2-over-(x^3+1)^2


3. Find the area, if it exists, of the region bounded by y= 1-over-square root of (3x-2)^3 and the lines x=6 and y=0

Answer
Hi, Donny,

You wrote:

Evaluate using integral table, indicate formula number and simplify as much as possible.


S=integral sign
1.
a) S square root sign over x^2(x squared) + 9dx


b) S ln 4-over-5 * xdx

c) S 2-over-5x(7x+2) * dx

d) S x^3e^-2x * dx

2. Determine whether the improper integral is convergent or divergent, and calculate its value if it is convergent (a & c)
a) S(infinity sign at the top; 0 at the bottom) 4e^-4x*dx

b) S(infinity sign at the top; 5 at the bottom) x^4*dx

c) S(infinity sign at the top; 1 at the bottom) 3x^2-over-(x^3+1)^2


3. Find the area, if it exists, of the region bounded by y= 1-over-square root of (3x-2)^3 and the lines x=6 and y=0
---------------------------------------
OK, this IS mathematics.  However, this is a lot of problems, so I will outline the solutions and you can work out the rest of the details. Keep in mind that this is not a site that does your homework for you.  It's intended to help you with the tough problems.  If all your problems are tough, maybe you are not ready for this subject.

(I'll also indicate the best notation to use for future problems.)

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

1a. This one takes a lot of work if you have to do it yourself.  However, since you said you could do it using an integral table, there is a result in the table (mine, anyway) that says:

{
| sqrt(a^2 + u^2) dx =
}

(u/2) sqrt(a^2 + u^2) + (a^2/2) ln | u + sqrt(a^2 + u^2) |

Now for  
{
| sqrt(x^2 + 9) dx
}

Identify:  u = x,  a = 3

and you are home.  If you can't do this, let me know:

1b. Is the following what the example says?
{     4
| ln(--) dx
}    5x

If so, use properties of logarithms to write it as:

{
| [ln 4 - ln 5 - ln x] dx
}    

The first two terms are constants -- no problem.  But the third term says:

{
| ln x dx
}    

and it is easy to find a formula for that.

1c. Again, I have to guess that you mean:

{      2
| ---------- dx
} 5x(7x + 2)

For this one, look for formulas involving  au + b  and you will find:

{      1
| ---------- dx
} u(au + b)

Simply take out the constant factor of 2/5, then identify:
u = x,  a = 7,  b = 2


1d. If this is the integral,
{
| x^3 e^(-2x) dx
}

then you look in the 'exponential' section and find:

{                  
| x^n e^(au) du =
}

1                n  {                  
--- u^n e^(au) - --- | x^(n-1) e^(au) du =
a                a  }

which is a kind of 'reduction formula'.  It reduces your 'x^3' example to an  x^2  example, then you use it again to reduce to an 'x' example then finally to an example with no x's, and then you can integrate it using a standard formula for  e^(au).  This will take you some time, of course, but it's routine.  

(Routine, he says -- all I need is a week off from work.)

-----------------------------------------
a) S(infinity sign at the top; 0 at the bottom) 4e^-4x*dx

b) S(infinity sign at the top; 5 at the bottom) x^4*dx

c) S(infinity sign at the top; 1 at the bottom) 3x^2-over-(x^3+1)^2

2a.  Does it say this?

{inf
|     4 e^(-4x) dx
}0
   e^(-4x)
= - -------  from  0 to inf.
     16

Now since  e^0 is 1 (nice and finite)  and
since  e^-infinity  is zero.  (Of course this means the limit of  

e^-b as b -> infinity)

The integral converges and is equal to 1/16.

2b.  I guess you figured out this one diverges.

{inf
|     x^4 dx
}5

The integral is  x^5/5 and clearly diverges at infinity.

2c.
{inf   3x^2
|    ----------- dx
}1   (x^3 + 1)^2

You will integrate this nicely and get:

something
---------  from  1 to infinity
x^3 + 1

At infinity, it will be zero, so the result is obtained at the

left-hand boundary.  I won't work it out, but when you do, get the

signs correct.  The integrand is certainly positive for all x, so

the result had better be a positive number.  If it isn't go back

and fix it.

-------------------------------------------------
For the third one:
3. Find the area, if it exists, of the region bounded by y=

1-over-square root of (3x-2)^3 and the lines x=6 and y=0

You basically want this integral

{6     
|  (3x-2)^-3/2     dx
}?   

Where does it start?  The function will be undefined if
3x - 2 <= 0  or
    x <= 2/3

So the left boundary will have to be  x = 2/3, and this integral is improper, because the integrand is undefined at x = 2/3.

Will it converge?  The integral is straightforward and you will have something like:

(3x - 2)^-1/2

in it.  But this will also be undefined at x = 2/3, so we are out of luck.