Why don't I get the correct answer when I set these two equations =

[rxh140630]

Homework Statement:

Find all x>0 for which $$\int_0^x [t]^{2} dt = 2(x-1) $$

Relevant Equations:

The notation [x] denotes the greatest integer less than or equal to x

In the question above it, the author (Apostol) states: $$\int_0^n [t]^{2} dt = \frac{n(n-1)(2n-1)}{6}$$

Why can't I set the two equations = and get the result?

2(x-1) = x(x-1)(2x-1)/6 => 12 = 2x^2 - x => 0 = x^2-(x/2) -6

using quadratic equation I get the wrong answer

 

[Infrared]

Apostol's formula is true when $n$ is a natural number. You aren't assuming that $x$ is a natural number, though.

 

[rxh140630]

Apostol's formula is true when $n$ is a natural number. You aren't assuming that $x$ is a natural number, though.

Ahh I see I see. Well how would you go about solving this problem without plug and chugging? Seems really hard

 

[Infrared]

Let $n$ be the floor of $x$. I would try breaking up the integral over the intervals $[0,n]$ and $[n,x]$ (for example, $\int_0^{7/2}=\int_0^3+\int_3^{7/2}$). You can use apostol's formula on the first piece, and directly evaluate the second.

 

[rxh140630]

Let $n$ be the floor of $x$. I would try breaking up the integral over the intervals $[0,n]$ and $[n,x]$ (for example, $\int_0^{7/2}=\int_0^3+\int_3^{7/2}$). You can use apostol's formula on the first piece, and directly evaluate the second.

Still seems like it would involve plug and chugging, I think I'm just going to have to skip this problem sadly.

 

[Infrared]

There's a (very) small amount of case work, but no blind 'plug and chug' needed.

 

[rxh140630]

There's a (very) small amount of case work, but no blind 'plug and chug' needed.

Guess I'll go back to it, lol.