The area of the region bounded by the line 3x - y + 2 = 0, the x-axis and the ordinates x = -1 and x = 1 is: Option 1: 4 sq.units Option 2: 13/3 sq.units Option 3: 1/6 sq.units Option 4: 25/6 sq.units

Answer: 13/3 sq.units. Solution: Here y = 3x+2, which crosses the x-axis at x = -2/3, so the curve lies below the axis on [-1,-2/3] and above on [-2/3,1]. Area =, _(-1)^(-2/3)(3x+2)dx, + _(-2/3)^1(3x+2)dx = 1/6 + 25/6 = 13/3 sq. units.

Calculus

Application of Integrals

Medium

common

The area of the region bounded by the line $3x - y + 2 = 0$ , the $x-axis$ and the ordinates $x = -1$ and $x = 1$ is:

4 sq.units

$\dfrac{13}{3}$ sq.units

$\dfrac{1}{6}$ sq.units

$\dfrac{25}{6}$ sq.units

✅ Correct Option: 2

Here $y = 3x+2$ , which crosses the $x$ -axis at $x = -\dfrac{2}{3}$ , so the curve lies below the axis on $\left[-1,-\dfrac{2}{3}\right]$ and above on $\left[-\dfrac{2}{3},1\right]$ .

Area $= \left|\int_{-1}^{-2/3}(3x+2)dx\right| + \int_{-2/3}^{1}(3x+2)dx = \dfrac{1}{6} + \dfrac{25}{6} = \dfrac{13}{3}$ sq. units.

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The area of the region bounded by the line 3x−y+2=03x - y + 2 = 03x−y+2=0, the x−axisx-axisx−axis and the ordinates x=−1x = -1x=−1 and x=1x = 1x=1 is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The area of the region bounded by the line $3x - y + 2 = 0$ , the $x-axis$ and the ordinates $x = -1$ and $x = 1$ is: