Let $[t]$ denote the largest integer less than or equal to $t$. If \[ \int_0^1 \left(\left[x^2\right] + \left\lfloor \frac{x^2}{2} \right\rfloor\right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}, \] where $a, b, c \in \mathbb{Z}$, then $a + b + c$ is equal to ________.
Correct Answer: 23
Approach Solution - 1
The problem asks to evaluate the definite integral \( \int_0^1 \left(\left[x^2\right] + \left\lfloor \frac{x^2}{2} \right\rfloor\right) dx \). However, for the interval \(x \in [0, 1)\), both \(x^2\) and \(x^2/2\) are less than 1, making both \([x^2]\) and \(\lfloor x^2/2 \rfloor\) equal to 0. This results in an integral value of 0, which contradicts the given form of the answer. A known version of this problem uses an upper limit of 3. Assuming the intended problem is to evaluate \( \int_0^3 \left(\left[x^2\right] + \left\lfloor \frac{x^2}{2} \right\rfloor\right) dx \), we proceed with this correction to find the value of \(a+b+c\).
Concept Used:
The greatest integer function, \([t]\) or \(\lfloor t \rfloor\), is a step function. The value of \(\lfloor f(x) \rfloor\) is constant between the points where the argument \(f(x)\) takes on integer values. To evaluate the definite integral of a step function, the interval of integration is broken down into sub-intervals at these points of discontinuity. The integral is then computed by summing the areas of the rectangles formed over each sub-interval, where the height of each rectangle is the constant value of the function in that sub-interval.
Step-by-Step Solution:
Step 1: Decompose the integral into two separate integrals.
Let \(I = \int_0^3 \left(\left[x^2\right] + \left\lfloor \frac{x^2}{2} \right\rfloor\right) dx\). We can split this as:
\[ I = \int_0^3 \left[x^2\right] dx + \int_0^3 \left\lfloor \frac{x^2}{2} \right\rfloor dx = I_1 + I_2 \]Step 2: Evaluate the first integral, \(I_1 = \int_0^3 [x^2] dx\).
The value of \([x^2]\) changes when \(x^2\) is an integer. For \(x \in [0, 3]\), \(x^2\) ranges from 0 to 9. The points of discontinuity are \(x = \sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots, \sqrt{8}, 3\). We split the integral at these points:
\[ I_1 = \int_0^1 0\,dx + \int_1^{\sqrt{2}} 1\,dx + \int_{\sqrt{2}}^{\sqrt{3}} 2\,dx + \int_{\sqrt{3}}^{2} 3\,dx + \int_{2}^{\sqrt{5}} 4\,dx + \int_{\sqrt{5}}^{\sqrt{6}} 5\,dx + \int_{\sqrt{6}}^{\sqrt{7}} 6\,dx + \int_{\sqrt{7}}^{\sqrt{8}} 7\,dx + \int_{\sqrt{8}}^3 8\,dx \]Step 3: Compute the value of \(I_1\).
\[ I_1 = 0 + (\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) + 4(\sqrt{5}-2) + 5(\sqrt{6}-\sqrt{5}) + 6(\sqrt{7}-\sqrt{6}) + 7(2\sqrt{2}-\sqrt{7}) + 8(3-2\sqrt{2}) \] \[ = \sqrt{2}-1 + 2\sqrt{3}-2\sqrt{2} + 6-3\sqrt{3} + 4\sqrt{5}-8 + 5\sqrt{6}-5\sqrt{5} + 6\sqrt{7}-6\sqrt{6} + 14\sqrt{2}-7\sqrt{7} + 24-16\sqrt{2} \]Grouping the terms:
\[ I_1 = (-1+6-8+24) + (1-2+14-16)\sqrt{2} + (2-3)\sqrt{3} + (4-5)\sqrt{5} + (5-6)\sqrt{6} + (6-7)\sqrt{7} \] \[ I_1 = 21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7} \]Step 4: Evaluate the second integral, \(I_2 = \int_0^3 \lfloor \frac{x^2}{2} \rfloor dx\).
The value of \(\lfloor x^2/2 \rfloor\) changes when \(x^2/2\) is an integer, i.e., when \(x^2 = 2k\). For \(x \in [0, 3]\), \(x^2/2\) ranges from 0 to 4.5. The points of discontinuity are \(x = \sqrt{2}, \sqrt{4}, \sqrt{6}, \sqrt{8}\). We split the integral accordingly:
\[ I_2 = \int_0^{\sqrt{2}} 0\,dx + \int_{\sqrt{2}}^2 1\,dx + \int_2^{\sqrt{6}} 2\,dx + \int_{\sqrt{6}}^{\sqrt{8}} 3\,dx + \int_{\sqrt{8}}^3 4\,dx \]Step 5: Compute the value of \(I_2\).
\[ I_2 = 0 + (2-\sqrt{2}) + 2(\sqrt{6}-2) + 3(2\sqrt{2}-\sqrt{6}) + 4(3-2\sqrt{2}) \] \[ = 2-\sqrt{2} + 2\sqrt{6}-4 + 6\sqrt{2}-3\sqrt{6} + 12-8\sqrt{2} \]Grouping the terms:
\[ I_2 = (2-4+12) + (-1+6-8)\sqrt{2} + (2-3)\sqrt{6} \] \[ I_2 = 10 - 3\sqrt{2} - \sqrt{6} \]Step 6: Calculate the total integral \(I = I_1 + I_2\).
\[ I = (21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}) + (10 - 3\sqrt{2} - \sqrt{6}) \] \[ I = (21+10) + (-3-3)\sqrt{2} - \sqrt{3} - \sqrt{5} + (-1-1)\sqrt{6} - \sqrt{7} \] \[ I = 31 - 6\sqrt{2} - \sqrt{3} - \sqrt{5} - 2\sqrt{6} - \sqrt{7} \]Step 7: Compare with the given expression to find \(a, b, c\).
The given form of the result is \(a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}\). Comparing this with our calculated value of \(I\):
\[ 31 - 6\sqrt{2} - \sqrt{3} - \sqrt{5} - 2\sqrt{6} - \sqrt{7} = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7} \]By comparing the coefficients, we get:
\[ a = 31 \] \[ b = -6 \] \[ c = -2 \]The values \(a, b, c\) are all integers.
Finally, we compute the value of \(a+b+c\):
\[ a+b+c = 31 + (-6) + (-2) = 31 - 8 = 23 \]The value of \(a+b+c\) is 23.
Approach Solution -2
Breaking the integral into intervals based on values of \( x \):
\[\int_0^3 \left( \lfloor x^2 \rfloor + \left\lfloor \frac{x^2}{2} \right\rfloor \right) dx\]
\[= \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{\sqrt{5}} 3 \, dx + \int_{\sqrt{5}}^{\sqrt{6}} 4 \, dx + \int_{\sqrt{6}}^{\sqrt{7}} 5 \, dx\]
\[+ \int_{\sqrt{7}}^{\sqrt{8}} 6 \, dx + \int_{\sqrt{8}}^3 7 \, dx\]
After evaluating each integral, we get:
\[a = 31, \quad b = -6, \quad c = -2\]
Thus,
\[a + b + c = 31 - 6 - 2 = 23\]
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