Question

Let [t] denote the greatest integer function. If \(\int\limits_0^{2.4}[x^2]dx=α+β√2+γ√3+δ√5,\) then α+β+γ +δ is equal to _____.

Hint:

To solve integrals with the greatest integer function, identify intervals where the function is constant and calculate the definite integral for each segment.

Answer 1

Correct Answer: 6

The greatest integer function \( [x^2] \) takes constant integer values over specific intervals of \(x\), so split the integral based on these intervals:

1. Intervals for \( [x^2] \):

• For \(0 \leq x < 1\), \(x^2 \in [0, 1)\), so \( [x^2] = 0 \).
• For \(1 \leq x < \sqrt{2}\), \(x^2 \in [1, 2)\), so \( [x^2] = 1 \).
• For \( \sqrt{2} \leq x < \sqrt{3} \), \(x^2 \in [2, 3)\), so \( [x^2] = 2 \).
• For \( \sqrt{3} \leq x < \sqrt{5} \), \(x^2 \in [3, 5)\), so \( [x^2] = 3 \).
• For \( \sqrt{5} \leq x < 2.4 \), \(x^2 \in [5, 2.4^2)\), so \( [x^2] = 4 \).

2. Evaluate each integral:

• \( \int_0^1 0 \, dx = 0. \)
• \( \int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1. \)
• \( \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2}). \)
• \( \int_{\sqrt{3}}^{\sqrt{5}} 3 \, dx = 3(\sqrt{5} - \sqrt{3}). \)
• \( \int_{\sqrt{5}}^{2.4} 4 \, dx = 4(2.4 - \sqrt{5}). \)

3. Combine all results:

\[ \int_0^{2.4} [x^2] dx = (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(\sqrt{5} - \sqrt{3}) + 4(2.4 - \sqrt{5}). \]

Simplify:

\[ = 9 - \sqrt{2} - \sqrt{3} - \sqrt{5}. \]

4. Match the format:

Compare with \( \alpha + \beta \sqrt{2} + \gamma \sqrt{3} + \delta \sqrt{5} \), so:

\[ \alpha = 9, \quad \beta = -1, \quad \gamma = -1, \quad \delta = -1. \]

5. Sum the coefficients:

\[ \alpha + \beta + \gamma + \delta = 9 - 1 - 1 - 1 = 6. \]

Final Answer:

\[ 6. \]