1
$\begingroup$

I = $\int _{0}^{3}\left( 1+x^{2}\right) d[ x]$

$a)12$

$b)17$

$c)15$

$d)19$

where $[x]$ is the greatest integer less than or equal to $x$,

shouldn't this integral be $0$ since $d[x]$ is 0 for all $x$ in the domain of the function. what is the correct explanation ??

$\endgroup$
1
  • $\begingroup$ Can we not use $\int f(x)dg(x)=\int f(x) g’(x)dx$? $\endgroup$ Commented Mar 19 at 13:17

2 Answers 2

2
$\begingroup$

This is the Riemann-Stieltjes integral of $(1+x^2)$ from $0$ to $3$ with respect to $g(x)=[x]$.

This integral can be evaluated like a Riemann sum, partitioning $[0,3]$ into intervals with tags and increasing the number of intervals. The kind of sum we're trying to evaluate is:

$$ \lim_{n\to \infty}\sum_{i=1}^{n}f(x_i)\>|g(x_i)-g(x_{i-1})| $$

Here, we'll use the rightmost point as the tag for definiteness, but it doesn't change the answer. Notice that the function $g$ is constant in each of the open intervals $(0,1)$, $(1,2)$, and $(2,3)$, so the factors on the right are only nonzero when we evaluate $g$ on two different intervals: when $x_i=1,2,3$. The sum then reduces to:

$$ f(1) + f(2) + f(3) = 2 + 5 + 10 = 17$$

Another way to think about this is, as the comments suggest, using the definition $\int f(x)dg(x)=\int f(x) g’(x)dx$. Here, we need $g$ to be differentiable, but if we allow some hand-wavy use of the delta function then $g'(x) = \delta(x-1) + \delta(x-2) + \delta(x-3) $ which also gives the same result.

$\endgroup$
0
$\begingroup$

If you think integration as sum of areas of rectangles under the function, it will be easy. Like breadth of rectangles generally is $dx$ $\to $ $0$ but here $d[x]$ is I think equals to $1$. So make rectangles of breadth $1$. enter image description here

Answer will be sum of all

$\sum f(x)d[x]$ that is

.$ f(0)d[x] + f(0+d[x])d[x] + f(0+2d[x])d[x]$.

So answer becomes $2×1 + 5×1 + 10×1 = 17$

I hope this might help you.

$\endgroup$
2

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .