Can a function equal zero if the integral of its product is zero?

[aaaa202]

Suppose you are given a function:

g(y) = ∫_a^bf(x,y)dx
And you are told f(x,c)=0. Does this then imply that:
g(c)=∫_a^bf(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)?

 

[Mark44]

Suppose you are given a function:

g(y) = ∫_a^bf(x,y)dx
And you are told f(x,c)=0. Does this then imply that:
g(c)=∫_a^bf(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)?

Your shortcut looks fine to me.

 

[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 canceled t?

 

[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.