Can anyone show me how to prove the following:
For $x\in \left [ n,n+1 \right ]$: \int_{n}^{n+1}\int_{n}^{x}\left | f'(t) \right |dtdx\leqslant \int_{n}^{n+1}\left | f'(t)\right |dt?
Can anyone show me how to prove the following:
For $x\in \left [ n,n+1 \right ]$: \int_{n}^{n+1}\int_{n}^{x}\left | f'(t) \right |dtdx\leqslant \int_{n}^{n+1}\left | f'(t)\right |dt?
Hint:
$\int_a^b \int_a^x g(t)\,dt\,dx=\int_a^b (b-x)g(x)\,dx$
Why is this true? To go from the LHS to RHS intuitively, consider the fact that for $t\in [a,b]$, the double integral is set up to count $g(t)$ with generalized "multiplicity," or weight, proportionate to the length of the interval $[t,b]$ (this is the interval of $x$ values for which $t\in [a,x]$). To go from the RHS to the LHS algebraically, consider by-parts with $u=x-a,dv=g(x)$.
Hint: The integral runs over the domain $n\leq t\leq x\leq n+1$. Interchanging the $x$ and $t$ integration, we have \int_{n}^{n+1}\int_{n}^{x}\left | f^{'}(t) \right |dt\,dx = \int_{n}^{n+1}\int_{t}^{n+1}\left | f^{'}(t) \right |dx\,dt. I am sure you can take it from here...
Since $n\leq x\leq n+1$: \int_n^x|f'(t)|dt\leq \int_n^{n+1}|f'(t)|dt and if we integrate it with respect to $x$ over $[n,n+1]$, we get the expected result.