# Finding definite integral of f(x) but not given a function?

• May 1st 2009, 02:24 PM
ceb0196
Finding definite integral of f(x) but not given a function?
I think my prof is trying to trip me up. Here is the question(sorry, I don't know the codes to make it in the right form):

Suppose: integrand with b=6 and a=1 of f(x) dx=9, integrand with b=4 and a=6 of f(x) dx=4, then find integrand with b=4 and a=1

does that make any sense? How do I solve this without being given a function??
• May 1st 2009, 02:27 PM
mr fantastic
Quote:

Originally Posted by ceb0196
I think my prof is trying to trip me up. Here is the question(sorry, I don't know the codes to make it in the right form):

Suppose: integrand with b=6 and a=1 of f(x) dx=9, integrand with b=4 and a=6 of f(x) dx=4, then find integrand with b=4 and a=1

does that make any sense? How do I solve this without being given a function??

You're given $\displaystyle \int_1^6 f(x) \, dx = 9$ and $\displaystyle \int_4^6 f(x) \, dx = 4$.

Note that $\displaystyle \int_1^6 f(x) \, dx = \int_1^4 f(x) \, dx + \int_4^6 f(x) \, dx \Rightarrow \int_1^4 f(x) \, dx = \int_1^6 f(x) \, dx - \int_4^6 f(x) \, dx$.
• May 1st 2009, 02:32 PM
Plato
Is this your question: $\displaystyle \int_1^6 f = 9\;\& \;\int_6^4 f = 4\; \Rightarrow \;\int_1^4 f = ?$?

If so, notice that $\displaystyle \int_4^6 f = - 4\;\& \;\int_1^4 f + \int_4^6 f = \int_1^6 f$.
• May 1st 2009, 02:35 PM
ceb0196
Yes, that is the question. Thanks to both for the start; I will try to work it out!
• May 1st 2009, 02:52 PM
ceb0196
I just can't figure out what I am missing. How do I solve without having a function that I can plug F(b) and F(a) into??
• May 1st 2009, 02:58 PM
mr fantastic
Quote:

Originally Posted by ceb0196
I just can't figure out what I am missing. How do I solve without having a function that I can plug F(b) and F(a) into??

Read posts #2 and #3 again. Read them carefully. Where is your trouble in plugging in the value of things that you know and solving for the thing that you don't know?
• May 1st 2009, 03:14 PM
ceb0196
Is it 13? It can't possibly be that simple...
• May 1st 2009, 03:26 PM
Plato
Quote:

Originally Posted by ceb0196
Is it 13? It can't possibly be that simple...

Oh yes, it is indeed that simple!
• May 1st 2009, 03:36 PM
ceb0196
Oh my goodness gracious. Thanks to all of you for helping me fumble through the last week of this class!