1. ## Integration Proofs

I'm really lost on these ones

(a) suppose f and g are continuous functions on [a,b] such that $\int_a^b f = \int_a^b g$. Prove that $\exists x \in [a,b]$ such that f(x)=g(x)

(b) suppose f is an integrable function on [a,b] and suppose that g is a function such that f(x)=g(x) except for finitely many $x \in [a,b]$. Show that g is integrable on [a,b] and $\int_a^b f = \int_a^b g$

2. Originally Posted by akolman
I'm really lost on these ones

(a) suppose f and g are continuous functions on [a,b] such that $\int_a^b f = \int_a^b g$. Prove that $\exists x \in [a,b]$ such that f(x)=g(x)
$\int_a^b f = \int_a^b g$

$\Rightarrow \int_a^b f - \int_a^b g = 0$

$\Rightarrow \int_a^b f-g = 0$

now what can you say?

(b) suppose f is an integrable function on [a,b] and suppose that g is a function such that f(x)=g(x) except for finitely many $x \in [a,b]$. Show that g is integrable on [a,b] and $\int_a^b f = \int_a^b g$
Suppose $g$ is the function zero. So that $f(x) = 0$ for all $x \in [a,b]$ except for finitely many points $y_1, \dots , y_N$. show that $\int_a^b f(x)~dx = 0$.

Once you do the above, apply the argument to the difference $f - g$, to get $\int_a^b f - g = 0 \implies \int_a^b f = \int_a^b g$ (use part (a))