CBSE Class 12 · 2016 · Area Under CurvesLong Answer

Using the method of integration, find the area of the triangular region whose vertices are

(2, - 2)

(4, 3)

and

(1, 2)

Answer: $\frac{13}{2}$ square units

Solution

Label

A (2, - 2)

B (4, 3)

C (1, 2)

.
Equation of

A B

2 y = 5 x - 14

; equation of

B C

3 y = x + 5

; equation of

C A

4 x + y = 6

.
Integrating with respect to

y

(convenient here), with

A B

on the right and

C A

B C

on the left:

A = \int_{- 2}^{3} \frac{2 y + 14}{5} d y - \int_{- 2}^{2} \frac{6 - y}{4} d y - \int_{2}^{3} (3 y - 5) d y = \frac{13}{2} .

(Equivalently, by the determinant check,

\frac{1}{2} ∣2 (3 - 2) + 4 (2 + 2) + 1 (- 2 - 3) ∣ = \frac{1}{2} (13) = \frac{13}{2}

.)
Hence the area of the triangular region is

\frac{13}{2}

square units.

CBSE Class 12 Board · 2016 · Subjective