Expert: Paul Klarreich - 11/13/2008

Question
I am ready to pull my hair out. I have been at this all day. I understand how to evaluate definite integrals when they are evaluated using numbers.  However in a problem like the one below, I just crumble.  Any help on understanding this concept would be really appreciated. That weird symbol under the 2sqrt was suppose to be the integral sign but it didnt copy very well.

2sqrtx
ò   x sin(6x^2)
0


Answer
Hi, Heather,

Here are two ideas for you:

1. Read some prior questions and answers on this site. There are a TON of them.  They won't answer your question but they will show you how to input math expressions.

2. This type of expression:

{72
|   f(t) dt
}31

is a number, which depends on the constants(the 31 and the 72), and on the particular function f.  When you are done, there won't be any t's in it.  That makes 't' something called a DUMMY VARIABLE.  

This expression:

{72
|   f(x) dx
}31

is EXACTLY THE SAME NUMBER AS THE FIRST ONE.

{something-with-x-in-it.
|   f(t) dt
}17

is a function of x.  You evaluate it by doing the usual stuff -- no magic about it.

1. find an antiderivative of  f(t).
2. Put something-with-x-in-it in place of t for the right boundary.
3. Put 17 in place of t for the left boundary.
4. Subtract.  You will get something with x's in it.  That's how it gets to be a function of x.

And, one more thing:

{x
|   f(z) dz
}17
and
{x
|   f(s) ds
}17
and
{x
|   f(u) du
}17

AND EVEN:

{x
|   f(x) dx
}17

are all exactly the same.  Yes, even the last one.  And in that one, the 'x' that is a boundary and the 'x' that is a variable of integration ARE NOT THE SAME THING.  Use different font faces, if you like, change the letter if you like.

Beginning to get the idea?


I think you wrote:


{2 sqrt(x)               << this x here is not the same as   
|   x sin(6x^2)  dx      << this x, which is a dummy variable.
}0                       << that's a zero??

Oh, and don't forget the dx.  You must have it. Don't be lazy.

Antidifferentiate:

{2 sqrt(x)
|   x sin(6x^2)  dx
}0

Do a standard u-subst:

u = 6x^2,   du = 12x dx,  dx = du/12x

{ << don't write this boundary when you have u's.
|   x sin(u)  du/12x
} << don't write this boundary, either.


{
| (1/12)   sin(u)  du
}

= - (1/12)cos u = -(1/12) cos (6x^2).

from  x = 0  to  x = sqrt(x)


-(1/12) [ cos (6(sqrt(x)^2) - (- cos (6(0)^2)) ]

-(1/12) [ cos (6x) - (- cos (0)) ]

-(1/12) [ cos (6x) + 1 ]

FINITO.