CBSE Class 12 · 2017 · Area Under CurvesLong Answer

Using integration, find the area of region bounded by the triangle whose vertices are

(- 2, 1)

(0, 4)

and

(2, 3)

Answer: $4$ square units

Solution

Label

A (- 2, 1)

B (0, 4)

C (2, 3)

.
Equation of

A B

y = \frac{3}{2} x + 4

; equation of

B C

y = 4 - \frac{x}{2}

; equation of

A C

y = \frac{1}{2} x + 2

.
With

A C

as the lower edge over

x \in [- 2, 2]

, upper edge

A B

for

- 2 \leq x \leq 0

then

B C

for

0 \leq x \leq 2

A = \int_{- 2}^{0} (\frac{3}{2} x + 4) d x + \int_{0}^{2} (4 - \frac{x}{2}) d x - \int_{- 2}^{2} (\frac{1}{2} x + 2) d x = 5 + 7 - 8 = 4.

Hence the area of the triangle is

4

square units.

CBSE Class 12 Board · 2017 · Subjective