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The definite integral of a function computes the area under the graph of its curve, allowing us to calculate areas and volumes that are not easily done using geometry alone.

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\[\bigg \lfloor \;10 \times \displaystyle \int_2^e \! \frac {1}{\ln(x)} \, \mathrm{d}x\;\bigg \rfloor = \ ? \]

**Details and Assumptions**:

- You may use the following approximations: \(e \approx 2.718\), and \(\ln(2) \approx 0.693\), and use the following graph of \(f(x)= \frac {1}{\ln(x)}\).

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\[\displaystyle \int_{0}^{12} \Bigl( \{x\}^2+\lfloor x \rfloor ^2\Bigr) \ dx = \ ? \]

**Details and assumptions**:

- Every \(x\in \mathbb{R}\) can be written as \(x=\lfloor x \rfloor + \{x\} \).
- As usual, \(\lfloor x \rfloor\) denotes greatest integer less than or equal to \(x\).
- \(\{x\} \) is the fractional part of \(x\).

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If \( \int_0 ^ {3} f(x) \, dx = 2 \), what is the value of

\[ \int_0^{3} \left( 3 - f(x) \right) \, dx ? \]

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