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MHTCET 15 May 2024 Shift 1 Paper

Question : 124 of 150

Marks: +1, -0

y=3 x+1, y=4 x+1

x=2

Solution:

Concept:

The area between two curves is found by integrating the difference of the functions over the given interval.

Explanation:

The lines

y = 3x + 1

and

y = 4x + 1

intersect at

x = 0

The region is bounded by these two lines and the vertical line

x = 2

For

0 \le x \le 2

, the upper curve is

y = 4x + 1

and the lower curve is

y = 3x + 1

Required area is:

A = \int_{0}^{2} \left[(4x + 1) - (3x + 1)\right] dx

A = \int_{0}^{2} x \, dx

A = \left[\frac{x^2}{2}\right]_{0}^{2}

A = \frac{4}{2} - 0 = 2 \text{ sq. units}

Answer:

The area is

2

sq. units.

Correct option: B. 2 sq. units.

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