We need to evaluate $I=\int_0^{2.4}[x^2] \, dx,$ where $[t]$ denotes the greatest integer less than or equal to $t$.

1. Break the interval according to values of $[x^2]$

The value of $[x^2]$ changes whenever $x^2$ crosses an integer.

For $0\le x\le 2.4$, we have $0\le x^2\le (2.4)^2=5.76.$ So possible values of $[x^2]$ are $0,1,2,3,4,5.$

Now:

• $[x^2]=0$ when $0\le x^2<1 \implies 0\le x<1$
• $[x^2]=1$ when $1\le x^2<2 \implies 1\le x<\sqrt2$
• $[x^2]=2$ when $2\le x^2<3 \implies \sqrt2\le x<\sqrt3$
• $[x^2]=3$ when $3\le x^2<4 \implies \sqrt3\le x<2$
• $[x^2]=4$ when $4\le x^2<5 \implies 2\le x<\sqrt5$
• $[x^2]=5$ when $5\le x^2<6 \implies \sqrt5\le x\le 2.4$

Thus, $I=\int_0^1 0\,dx+\int_1^{\sqrt2}1\,dx+\int_{\sqrt2}^{\sqrt3}2\,dx+\int_{\sqrt3}^{2}3\,dx+\int_2^{\sqrt5}4\,dx+\int_{\sqrt5}^{2.4}5\,dx.$

2. Compute each part

$\int_0^1 0\,dx=0$

$\int_1^{\sqrt2}1\,dx=\sqrt2-1$

$\int_{\sqrt2}^{\sqrt3}2\,dx=2(\sqrt3-\sqrt2)$

$\int_{\sqrt3}^{2}3\,dx=3(2-\sqrt3)=6-3\sqrt3$

$\int_2^{\sqrt5}4\,dx=4(\sqrt5-2)=4\sqrt5-8$

$\int_{\sqrt5}^{2.4}5\,dx=5(2.4-\sqrt5)=12-5\sqrt5$

3. Add them

So, $I=(\sqrt2-1)+2(\sqrt3-\sqrt2)+(6-3\sqrt3)+(4\sqrt5-8)+(12-5\sqrt5).$

Expand: $I=\sqrt2-1+2\sqrt3-2\sqrt2+6-3\sqrt3+4\sqrt5-8+12-5\sqrt5.$

Combine like terms:

• Constant terms: $-1+6-8+12=9$
• $\sqrt2$ terms: $\sqrt2-2\sqrt2=-\sqrt2$
• $\sqrt3$ terms: $2\sqrt3-3\sqrt3=-\sqrt3$
• $\sqrt5$ terms: $4\sqrt5-5\sqrt5=-\sqrt5$

Hence, $I=9-\sqrt2-\sqrt3-\sqrt5.$

So, $\alpha=9,\quad \beta=-1,\quad \gamma=-1,\quad \delta=-1.$

Therefore, $\alpha+\beta+\gamma+\delta=9-1-1-1=6.$

4. Comparison with stored answer

Derived answer = $6$. Stored correct answer = $6$. They match.