# Function given as integral

1. Oct 28, 2013

### aaaa202

Suppose you are given a function:

g(y) = ∫abf(x,y)dx
And you are told f(x,c)=0. Does this then imply that:
g(c)=∫abf(x,c)dx=∫0dx = 0
Or are you supposed to calculate g(y) from the integral first and then plug in c to find g(c)?

2. Oct 28, 2013

### Staff: Mentor

Your shortcut looks fine to me.

3. Oct 29, 2013

### aaaa202

But what if the you had something like:
∫f(x,t)tdx
And f(x,a)=0. But the integration would yield something that cancelled t?

Last edited: Oct 29, 2013
4. Oct 29, 2013

### arildno

No, not unless the product function with f is sufficiently nasty.
REMEMBER that you are basically adding together the areas of tiny rectangles of height f(x,a)*a in your new case.

If f(x,a)=0 for all those rectangles, then the sum is zero.

IF, however, you had something under the integral sign:
f(x,t)/(t-a), then even though f(x,a)=0, you cannot conclude that f(x,a)/(a-a)=0, or is even defined.