How do you show
$$ \int ^{b}_{a} f(x) \,dx \leq \int ^{b}_{a} g(x) \,dx$$
if you know f(x) and g(x) are continuous over [a,b] and $$f(x) \leq g(x)$$ for $${a \leq x \leq b}$$
Here is the way I would solve it.
Show $$ \int ^{b}_{a} f(x) \, dx \leq \int ^{b}_{a} g(x) \, dx$$ Show $$ \int ^{b}_{a} f(x) \, dx - \int ^{b}_{a} g(x) \, dx \leq 0$$ Show $$ \int ^{b}_{a} (f(x) - g(x)) \, dx \leq 0$$
Since we know $$f(x) \leq g(x)$$ it follows that $$f(x) - g(x) \leq 0$$
which means:
Show $$ \int ^{b}_{a} \text{(something less than or equal to 0)} \, dx \leq 0$$
But obviously, if you integrate something less than or equal to $0$ you are going to end up with something less than or equal to 0 [as was to be shown].
Is this a correct proof ?
Well, lets pickup from th moment you've got $$\int^b_a(f(x)-g(x))\,dx\leq 0$$
To show this let's consider a function $H(x)$ defined as follows
$$H(x)=\int^x_a(f(t)-g(t))\,dt.$$
It's obvious that $H(a)=0$. We know that a decreasing function is a function, which has negative derivative, but
$$H^\prime(x)=f(x)-g(x)\leq 0$$
This means $H(x)$ is decerasing on $[a,b]$ and subsequently $0=H(a)\geq H(x),\,\forall x\in[a,b]$.