First slide

Evaluation of definite integrals

Question

The value of $\int_{- 1}^{3} {| x - 2 | + [x]} d x$ , where $[x]$ denotes the greatest integer less than or equal to x is

Easy

A

5
B

7
C

4
D

3

Solution

$\int_{- 1}^{3} {| x - 2 | + [x]} d x = \int_{- 1}^{0} {| x - 2 | + [x]} d x + \begin{aligned} \int_{0}^{1} {| x - 2 | + [x]} d x + \int_{1}^{2} {∣ x - 2 ∣ + [x]} d x \\ + \int_{2}^{3} {| x - 2 | + [x]} d x \end{aligned}$
$\begin{aligned} = \int_{- 1}^{0} (2 - x - 1) d x + \int_{0}^{1} (2 - x + 0) d x + \\ \int_{1}^{2} (2 - x + 1) d x + \int_{2}^{3} (x - 2 + 2) d x \end{aligned}$
$\begin{array}{l} = x - {\frac{x^{2}}{2}|}_{- 1}^{0} + 2 x - {\frac{x^{2}}{2}|}_{0}^{1} + 3 x - {\frac{x^{2}}{2}|}_{1}^{2} + {\frac{x^{2}}{2}|}_{2}^{3} \\ = - (- 1 - \frac{1}{2}) + (2 - \frac{1}{2}) + (6 - 2) - (3 - \frac{1}{2}) + \frac{9}{2} - 2 = 7 \end{array}$

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