CUET Mathematics 2024 - The area of the region bounded by the lines x+2 y=12, x=2, x=6 and x-axis is : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Application of Integrals

Easy

The area of the region bounded by the lines $x+2 y=12, x=2, x=6$ and $x$ -axis is :

34

sq units

20

sq units

24

sq units

16

sq units

✅ Correct Option: 4

Identifying the region by finding the intersection points:

- The

x

-axis corresponds to

y=0

\newline

- When

y=0

x+2y=12

, we get

x=12

\newline

- Our region is bounded by

x=2

and

x=6

, which are vertical lines

Expressing

y

in terms of

x

for the sloped boundary:

From

x+2y=12

, I can solve for

y

2y = 12-x

y = \dfrac{12-x}{2} = 6-\dfrac{x}{2}

Finding intersection points to define the region completely:

x=2

y = 6-\dfrac{2}{2} = 5

x=6

y = 6-\dfrac{6}{2} = 3

Calculating the area using integration:

\text{Area} = \int_{2}^{6} (6-\dfrac{x}{2}) \, dx

\text{Area} = [6x - \dfrac{x^2}{4}]_{2}^{6}

\text{Area} = \left(6(6) - \dfrac{6^2}{4}\right) - \left(6(2) - \dfrac{2^2}{4}\right)

\text{Area} = \left(36 - \dfrac{36}{4}\right) - \left(12 - \dfrac{4}{4}\right)

\text{Area} = \left(36 - 9\right) - \left(12 - 1\right)

\text{Area} = 27 - 11 = 16

Alternatively, using the trapezoid formula:

\text{Area} = \dfrac{1}{2} \times \text{width} \times (\text{height}_1 + \text{height}_2) = \dfrac{1}{2} \times 4 \times (5 + 3) = 16

Therefore, the area of the region is 16 square units.