How to integrate this expression? Please help me out.

Question

How to integrate $$\int_{4/3}^{17/10}[x^{2}+3x+\frac{5}{3}]dx$$; where [.] denotes Greatest Integer Function. I am trying this question by factorizing $x^{2}+3x+\frac{5}{3}$. So, it can be factorized as $(x+\frac{3}{2})^{2}-\frac{7}{12}$. But what to do after factorizing? I can't proceed further. Please help me out.

Answer 1

Hint

In the range of integration, $$x^2+3x+\frac 53$$ has integer values $\{8,9,10\}$.

So, solve the quadratic $$x^2+3x+\frac 53=k \qquad \text{for } k=8,9,10$$ to obtain the bounds of integration.

Answer 2

I figured out a solution that's slightly calculative, but doesn't involve the factorization approach..

Here's what I did:

1. Since the inner expression is a quadratic, -b/a will provide us the minimum value which is at -3/2 as the x-coordinate or the input.

2. 4/3 which is the lower limit is ahead of -3/2 and the function increases after -3/2, on putting 4/3 into the quadratic we get a decimal number slightly more than 7 which on using the greatest integer would give the output as 7.

3. The entire function (along with the greatest integer) keeps giving the value as 7 as long as the quadratic produces an output of 8, lets say at a point q. The integration can now be broken from 4/3 to q with 7dx to be integrated which can be integrated into 7x with limits from 4/3 to q. The value of q comes out to be 1.43.

4. The output of the quadratic at the upper limit (17/10) comes out to be around 9.65 which the greatest integer will floor to 9. Now we find the value for which the quadratic gives an output of 9 which comes out to be 1.596.

5. Now from 1.43 to 1.596 the integral would be 8dx which can be integrated to give 8x with limits 1.43 to 1.596.

6. Finally, from 1.596 to 17/10, the entire function gives 9 as the output which upon integration gives 9x with limits 1.596 to 17/10.

To sum up: Break the limits of integration at points where the function provides an integral value, the entire floor input now behaves as an integer in that limit which is easy to solve.

Answer 3

Let $f(x)=\lfloor x^{2}+3x+\frac{5}{3}\rfloor$. Then as Claude Leibovici suggested we solve $$x^2+3x+5/3=k$$ where $k=8,9$ (note that $f(4/3)>7$ and $f(17/10)<10$) for boundries and split the integral as follows: $$\int_{4/3}^{17/10}f(x)dx=\int_{4/3}^{\sqrt{103/12}-3/2}7dx+\int_{\sqrt{103/12}-3/2}^{\sqrt{115/12}-3/2}8dx+\int_{\sqrt{115/12}-3/2}^{17/10}9dx$$