1. ## integration theorem!

i need some help or understanding or a book or material that explains this type of problem

find the value of

$\int_{0.3}^{2.6} x^3 d(x + [x] )$

thanks

2. ## Re: integration theorem!

That is an example of a "Stieljes" integral. The original "Riemann" integral, $\int_a^b f(x)dx$, divides the x-axis into sub-intervals, $[x_i, x_{i+1}]$, and then measures the "area" by $f(x)\Delta x$ where $\Delta x$ is just the length, $x_{i+1}- x_i$. The "Stieljes" integral, $\int_a^b f(x)d\alpha(x)$, generalizes that by measuring each sub-interval by $\alpha(x_{i+1})- \alpha(x_i)$ where $\alpha(x)$ can be any increasing function of x. If $\alpha(x)$ happens to be differentiable the Stieljes integral reduces to a Riemann integral: $\int_a^b f(x)d\alpha= \int_a^b f(x)\alpha'(x)dx$.

But $\alpha(x)$ does not have to be differentiable or even continuous. One application of the the Stieljes integral is to be able to write a sum as an integral: if $\alpha(x)= \lfloor x\rfloor$ then $\alpha(x_{i+1})- \alpha(x_i)$ is non-zero only for $x_{i+1}$ and $x_i$ to be on different sides of an integer: $\int_0^5 f(x)d\lfloor x\rfloor= f(0)+ f(1)+ f(2)+ f(3)+ f(4)+ f(5)$.

That, together with the fact that
$\int_{0.3}^{2.6} x^3d(x+ \lfloor x\rfloor)= \int_{0.3}^{2.6}x^3 dx+ \int_{0.3}^{2.6} x^3 d\lfloor x\rfloor$
should be enough for you to do this problem.

(By the way, I have interpreted your "[x]" as the floor function. Latex has "lfloor" and "rfloor" to get $\lfloor x\rfloor$.

3. ## Re: integration theorem!

do u carry out normal integration on

$\int_{0.3}^{2.6} x^3 dx$

thanks

4. ## Re: integration theorem!

Yes, of course.