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$\begingroup$

In example they changed.

$\int_0^1 [5x] dx = \int_0^{\frac 15} [5x] dx + \int_{\frac 15}^{\frac 25} [5x] dx + \int_{\frac 1 25}^{\frac 35} [5x] dx + \int_{\frac 35}^{\frac 45} [5x] dx + \int_{\frac 45}^1 [5x] dx$

$= \int_0^{\frac 15} 0 dx + \int_{\frac 15}^{\frac 1 25} 1 dx + \int_{\frac 25}^{\frac 35} 2 dx + \int_{\frac 35}^{\frac 45} 3dx + \int_{\frac 45}^1 4 dx$

In another one -

$\int_{0.5}^{3.5} [x] dx = \int_{0.5}^{1} [x] dx + \int_{1}^{2} [x] dx + \int_{2}^{3} [x] dx + \int_{3}^{3.5} [x] dx $

$= \int_{0.5}^{1} 0 dx + \int_{1}^{2} 1 dx + \int_{2}^{3} 2 dx + \int_{3}^{3.5} 3 dx $

So I have two questions -

1.) What this [ ] means in this is question? And any method to solve them?

2.) How they change limits in these type of questions? As you can see in above examples.

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    Are you familiar with the greatest integer function? Your Question 1 seems to indicate that you don't know, yet the title of the question shows you do... Also, do you know properties of definite integrals? – 2017-02-12
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    @Tim Thayer yes I know some properties. – 2017-02-12

1 Answers 1

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$[x]$ denotes the greatest integer function (Otherwise known as the floor function), which takes the largest integer less than or equal to $x$.

Since you cannot integrate $\int [x]~dx$ on its own, you must separate it into several functions which you can integrate.

For example, with the first example: $$\int_0^1 [5x]~dx$$ It is known that $[5x]=0$ when $0\leq x <0.2$, and is known that $[5x]=1$ as soon as $x=0.2$ ($0.2\leq x<0.4$) and so on as shown on the graph below. Since constants can be integrated, we can convert the function as they've done:

enter image description here

$$\int_0^1 [5x]~dx=\int_0^{0.2} 0~dx+\int_{0.2}^{0.4} 1~dx+\int_{0.4}^{0.6} 2~dx+\int_{0.6}^{0.8} 3~dx+\int_{0.8}^{1} 4~dx$$