∫−11 x2+x−3dx where {x } denotes the fractional part of x, is equal to

A
$\frac{1}{3} (1 + 3 \sqrt{5})$
B
$\frac{1}{6} (1 + 3 \sqrt{5})$
C
$\frac{1}{3} (3 \sqrt{5} - 1)$
D
$\frac{1}{6} (3 \sqrt{5} - 1)$

Solution:

$\begin{array}{l} \int_{- 1}^{1} \{x^{2} + x - 3\} dx = \int_{- 1}^{1} \{x^{2} + x\} dx \\ = \int_{- 1}^{1} (x^{2} + x - [x^{2} + x]) dx \\ = {[\frac{x^{3}}{3} + \frac{x^{2}}{2}]}_{- 1}^{1} - \int_{- 1}^{0} [x^{2} + x] dx - \int_{0}^{\frac{\sqrt{5} - 1}{2}} [x^{2} + x] dx - \int_{(\sqrt{5} - 1) 2}^{1} [x^{2} + x] dx \\ = \frac{2}{3} + 1 - 0 - 1 (1 - \frac{\sqrt{5} - 1}{2}) \\ = \frac{2}{3} + \frac{\sqrt{5} - 1}{2} \\ = \frac{1}{6} (1 + 3 \sqrt{5}) \end{array}$

$\int_{- 1}^{1} \{x^{2} + x - 3\} dx$ where {x } denotes the fractional part of x, is equal to

Solution:

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