CBSE Class 12 · 2011 · Area Under CurvesLong Answer

Sketch the graph of

y = ∣ x + 3∣

and evaluate the area under the curve

y = ∣ x + 3∣

above

x

-axis and between

x = - 6

x = 0

Answer: $9$ square units

Solution

The graph of

y = ∣ x + 3∣

is a V with vertex at

(- 3, 0)

: it is

y = - (x + 3)

for

x \leq - 3

and

y = x + 3

for

x \geq - 3

.
On

[- 6, 0]

the area under the curve splits at

x = - 3

A = \int_{- 6}^{- 3} - (x + 3) d x + \int_{- 3}^{0} (x + 3) d x .

Each piece is a right triangle of legs

3

and

3

, so each has area

\frac{1}{2} \cdot 3 \cdot 3 = \frac{9}{2}

A = \frac{9}{2} + \frac{9}{2} = 9 square units.

CBSE Class 12 Board · 2011 · Subjective