Question

If [x] denotes the greatest integer function then $\int_0^5 x^2[x]dx =$

Answer 1

$\frac{400}{3}$

Answer 2

Solution Explanation:

1. Write the integral as a sum over intervals where $[x]$ is constant:

   $$\int_0^5 x^2 \, [x] \,dx = \sum_{n=0}^{4} n \int_n^{n+1} x^2 \, dx.$$

2. Evaluate each integral:

   $$\int_n^{n+1} x^2 \, dx = \frac{(n+1)^3 - n^3}{3}.$$

3. Compute for $n = 1,2,3,4$ (for $n = 0$, the term is 0):

   • For $n=1$: $1 \cdot \frac{2^3-1^3}{3} = \frac{7}{3}$
   • For $n=2$: $2 \cdot \frac{3^3-2^3}{3} = 2\cdot\frac{27-8}{3} = \frac{38}{3}$
   • For $n=3$: $3 \cdot \frac{4^3-3^3}{3}= 4^3-3^3 = 64-27= 37$
   • For $n=4$: $4 \cdot \frac{5^3-4^3}{3} = 4\cdot\frac{125-64}{3} = \frac{244}{3}$

4. Sum the results:

   $$\frac{7}{3}+\frac{38}{3}+37+\frac{244}{3} = \frac{(7+38+244)}{3}+37 = \frac{289}{3}+37.$$

   Convert $37$ to thirds: $37=\frac{111}{3}$

   $$\frac{289+111}{3} = \frac{400}{3}.$$