Definite Integration
General
Grade 12
Question:
<p>If $f(x) = \begin{cases} 3[x]-5\frac{|x|}{x}, & x \neq 0 \\ 2, & x = 0 \end{cases}$ then $\int_{-3/2}^{2} f(x) dx$ is equal to ([.] denotes the greatest integer function)</p>
<p>-\frac{11}{2}</p>
<p>-\frac{7}{2}</p>
<p>-6</p>
<p>-\frac{17}{2}</p>
Step-by-Step Solution
Key Concept: General
$3[x] - 5\frac{|x|}{x} = 3[x] - 5$ if $x > 0$<br>$= 3[x] + 5$, if $x < 0$<br>$\Rightarrow \int_{-3/2}^{2} f(x) dx = \int_{-3/2}^{-1} (-1) dx + \int_{-1}^{0} (2) dx + \int_{0}^{1} (-5) dx + \int_{1}^{2} (-2) dx$<br>$= -1\left(-1 + \frac{3}{2}\right) + 2(1) + 1(-5) + (-2) = -\frac{1}{2} + 2 - 5 - 2 = -\frac{11}{2}$
Correct Answer: A