CBSE Class 12 · 2023 · Area Under CurvesLong Answer

Using integration, find the area of the triangle whose vertices are

(- 1, 1)

(0, 5)

and

(3, 2)

Answer: $\frac{15}{2}$ square units

Solution

Label

A (- 1, 1)

B (0, 5)

C (3, 2)

.
Equation of

A B

: slope

= \frac{5 - 1}{0 - ( - 1 )} = 4

, so

y = 4 x + 5

.
Equation of

B C

: slope

= \frac{2 - 5}{3 - 0} = - 1

, so

y = 5 - x

.
Equation of

A C

: slope

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

, so

y = \frac{x + 5}{4}

.
With

A C

as the lower edge over

x \in [- 1, 3]

, and the upper edge

A B

for

- 1 \leq x \leq 0

then

B C

for

0 \leq x \leq 3

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

= 3 + \frac{21}{2} - 6 = \frac{15}{2} .

Hence the area of the triangle is

\frac{15}{2}

square units.

CBSE Class 12 Board · 2023 · Subjective