# Thread: Given an integral, find the others

1. ## Given an integral, find the others

Given 05f(x)dx=10 and 57f(x)dx=3, (by the way, that means the intervals for the first one are 0 and 5, and the 2nd one is 5 and 7)
Find...
(a)07f(x)dx and (b)05f(x)dx

I wasn't sure how to do this, because there were no variables, so I thought f(x)dx=3, but then the intervals are different.
So for like the first one, I know that the intervals are [0, 7]
Then what we're integrating is f(x)dx

But I'm a bit confused on how to start off with that.
I'm used to integrating when given an equation, but this one is giving the function.

Can someone explain to me?
Thanks.

2. ## Re: Given an integral, find the others

you're supposed to use THIS theorem: for a < b < c,

$\int_a^b f(x)\ dx + \int_b^c f(x)\ dx \ = \int_a^c f(x)\ dx$

3. ## Re: Given an integral, find the others

Originally Posted by Deveno
you're supposed to use THIS theorem: for a < b < c,

$\int_a^b f(x)\ dx + \int_b^c f(x)\ dx \ = \int_a^c f(x)\ dx$
Oh!
I see, thanks!
So I'm a bit confused, but making sure, that's the Additive Interval Property right?

And just wondering, do I just give the 2 intervals that were given, to find the solutions to (a) and (b)?
Like what would be the variable for a, b, c, and d.

Would it be like this?
05f(x)dx+57f(x)dx

That would become what (a) is like.
But after that, what would you do, if that was right?
By the way, the answer is 13 in the book.

4. ## Re: Given an integral, find the others

Surely you can see that when you apply the additive property that 10 + 3 = 13...

5. ## Re: Given an integral, find the others

ok, look, definite integrals and indefinite integrals are VERY different.

indefinite integrals: plug in a function, get another function (plus a constant):

$\int f(x)\ dx = F(x) + C$

definite integrals: plug in a function, get a NUMBER:

$\int_a^b f(x)\ dx = F(b) - F(a)$

you're already given "the numbers", you don't NEED the function F (or f).