# An integral question which i have never encountered before

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 ??

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

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.

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 $$1$$.

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 $$2×1 + 5×1 + 10×1 = 17$$