What is \int_{\sqrt{2}}^{2}f(x)\,dx equal to?
Consider the following for the two (02) items that follow : Let f(x)=[x^{2}] where [~] is the greatest integer function.
- A. 6-\sqrt{3}-2\sqrt{2} ✓
- B. 6-\sqrt{3}-\sqrt{2}
- C. 6-\sqrt{3}+2\sqrt{2}
- D. 6+\sqrt{3}-2\sqrt{2}
Correct Answer: A. 6-\sqrt{3}-2\sqrt{2}
Explanation
Split the integral at values where x^2 is an integer. \int_{\sqrt{2}}^{2} [x^2] \, dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{\sqrt{4}} 3 \, dx. This gives 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) = 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 6 - \sqrt{3} - 2\sqrt{2}.
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