Area Under Curves — CBSE Class 12 · 2020 Board Question with Solution

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

(2, - 2)

(4, 5)

and

(6, 2)

Answer: $10$ square units

Solution

Label

A (2, - 2)

B (4, 5)

C (6, 2)

.
It is convenient to integrate with respect to

y

. The sides are

A B : x = \frac{2}{7} (y + 9)

B C : x = - \frac{2}{3} (y - 11)

and

A C : x = y + 4

Area = \int_{- 2}^{2} (y + 4) d y - \frac{2}{3} \int_{2}^{5} (y - 11) d y - \frac{2}{7} \int_{- 2}^{5} (y + 9) d y

= \frac{1}{2} [(y + 4)^{2}]_{- 2}^{2} - \frac{1}{3} [(y - 11)^{2}]_{2}^{5} - \frac{1}{7} [(y + 9)^{2}]_{- 2}^{5} .

Carrying out the integration (with

A C

on the left,

A B

and

B C

on the right) gives

A = 16 + 15 - 21 = 10.

(Check by determinant:

\frac{1}{2} ∣2 (5 - 2) + 4 (2 + 2) + 6 (- 2 - 5) ∣ = \frac{1}{2} ∣6 + 16 - 42∣ = 10

.)
Hence the area of the triangle is

10

square units.

CBSE Class 12 Board · 2020 · Subjective