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Question

Mathematics Question on integral

Let [t][t] denote the largest integer less than or equal to tt. If 01([x2]+x22)dx=a+b235+c67,\int_0^1 \left(\left[x^2\right] + \left\lfloor \frac{x^2}{2} \right\rfloor\right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}, where a,b,cZa, b, c \in \mathbb{Z}, then a+b+ca + b + c is equal to ________.

Answer

Breaking the integral into intervals based on values of xx:
03(x2+x22)dx\int_0^3 \left( \lfloor x^2 \rfloor + \left\lfloor \frac{x^2}{2} \right\rfloor \right) dx
=010dx+121dx+232dx+353dx+564dx+675dx= \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{\sqrt{5}} 3 \, dx + \int_{\sqrt{5}}^{\sqrt{6}} 4 \, dx + \int_{\sqrt{6}}^{\sqrt{7}} 5 \, dx
+786dx+837dx+ \int_{\sqrt{7}}^{\sqrt{8}} 6 \, dx + \int_{\sqrt{8}}^3 7 \, dx
After evaluating each integral, we get:
a=31,b=6,c=2a = 31, \quad b = -6, \quad c = -2
Thus,
a+b+c=3162=23a + b + c = 31 - 6 - 2 = 23