First slide

Methods of integration

Question

$\int_{1}^{2} [x^{3} - 1] dx$ where [.] denotes the greatest integer function, is equal to

Difficult

A

6
B

12
C

15
D

None of these

Solution

$\begin{array}{l} 1 \leq x \leq 2 \Rightarrow 1 \leq x^{3} \leq 8 \Rightarrow 0 \leq x^{3} - 1 \leq 7 \\ I = \int_{1}^{2} [x^{3} - 1] dx = \int_{1}^{2^{1 / 3}} [x^{3} - 1] dx + \int_{2^{1 / 3}}^{3^{1 / 3}} [x^{3} - 1] dx + \dots + \int_{7^{1 / 3}}^{2} [x^{3} - 1] dx \end{array}$
now, $x \in (1, 2^{\frac{1}{3}}), then x^{3} \in [1, 2) or [x^{3} - 1] = 0$ and so on
therefore,
$\begin{array}{l} I = \int_{1}^{2^{1 / 3}} 0 dx + \int_{2^{1 / 3}}^{3^{1 / 3}} 1 \cdot dx + \dots \int_{7^{1 / 3}}^{2} 6 dx \\ \begin{array}{ll} = & [3^{1 / 3} - 2^{1 / 3}] + 2 [4^{1 / 3} - 3^{1 / 3}] + 3 [5^{1 / 3} - 4^{1 / 3}] + 4 [6^{1 / 3} - 4^{1 / 3}] + 5 [6^{1 / 3} - 5^{1 / 3}] + 6 [2^{1 / 3} - 7^{1 / 3}] \end{array} \\ = 12 - [7^{1 / 3} + 6^{1 / 3} + 5^{1 / 3} + 4^{1 / 3} + 3^{1 / 3} + 2^{1 / 3}] \end{array}$

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