Question
Sketch the graph y = |x + 3|. Evaluate $\int\limits_{-6}^{0}|\text{x}-3|\text{dx} $ . What does this value of the integral represent on the graph.
✓
Answer
we have,y = |x + 3| intesect x = 0 and x = -6 at (0, 3) and (-6, 3)
Now,
y = |x + 3|
= x + 3
integration represents the area enclosded by the graph from x = -6 to x = 0
$\text{A}=\int\limits_{-6}^{0}|\text{y}|\text{dx} $
$=\int\limits_{-6}^{-3}|\text{x}|\text{dx}+\int\limits_{-3}^{0}|\text{y}|\text{dx} $
$=\int\limits_{-6}^{-3}-(\text{x}+3)\text{dx}+\int\limits_{-3}^{0}(\text{x}+3)\text{dx} $
$=-\Big[\frac{\text{x}^{2}}{2}+3\text{x}\Big]^{-3}_{-6}+\Big[\frac{\text{x}^{2}}{2}+3\text{x}\Big]^{-3}_{0}$
$=-\Big[ \frac{9}{2}-9-\frac{36}{2}+18\Big]+\Big[0+0-\frac{9}{2}+9\Big]$
$=9 \ \text{sq.}\ \text{units}$
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