CBSE Class 12 · 2014 · Area Under CurvesLong Answer

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

(- 1, 2)

(1, 5)

and

(3, 4)

Answer: $4$ square units

Solution

Label

A (- 1, 2)

B (1, 5)

C (3, 4)

.
Equation of

A B

y = \frac{3 x + 7}{2}

; equation of

B C

y = \frac{11 - x}{2}

; equation of

A C

y = \frac{x + 5}{2}

.
With

A C

as the lower edge over

x \in [- 1, 3]

, upper edge

A B

for

- 1 \leq x \leq 1

then

B C

for

1 \leq x \leq 3

A = \int_{- 1}^{1} \frac{3 x + 7}{2} d x + \int_{1}^{3} \frac{11 - x}{2} d x - \int_{- 1}^{3} \frac{x + 5}{2} d x .

= \frac{1}{2} [14 + 22 - 28] = \frac{1}{2} (8) = 4.

Hence the area of the triangle is

4

square units.

CBSE Class 12 Board · 2014 · Subjective