The area of the region bounded by the line 3x-y+2=0 , the x-axis and…

The correct answer is Option (2) → \( \displaystyle\frac{13}{3} \) sq.units

The line is y = 3x + 2. It cuts the x-axis at \( x=-\frac{2}{3} \), which lies between −1 and 1, so the region is partly below and partly above the x-axis. Area is the integral of |y|, so the integral is split at \( x=-\frac{2}{3} \).

    \( \displaystyle \text{Area}=\int_{-1}^{-2/3}\left|3x+2\right|dx+\int_{-2/3}^{1}(3x+2)\,dx \)

    \( \displaystyle \int_{-1}^{-2/3}-(3x+2)\,dx=-\left[\frac{3x^2}{2}+2x\right]_{-1}^{-2/3} \)

    \( \displaystyle =-\left[\left(\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{2}-2\right)\right] \)

    \( \displaystyle =-\left[-\frac{2}{3}+\frac{1}{2}\right] \)

    \( \displaystyle =\frac{1}{6} \)

    \( \displaystyle \int_{-2/3}^{1}(3x+2)\,dx=\left[\frac{3x^2}{2}+2x\right]_{-2/3}^{1} \)

    \( \displaystyle =\left(\frac{3}{2}+2\right)-\left(-\frac{2}{3}\right) \)

    \( \displaystyle =\frac{7}{2}+\frac{2}{3} \)

    \( \displaystyle =\frac{25}{6} \)

    \( \displaystyle \text{Area}=\frac{1}{6}+\frac{25}{6} \)

    \( \displaystyle =\frac{13}{3}\ \text{sq. units} \)

Shortcut: Both parts are triangles with a vertex at \( \left(-\frac{2}{3},0\right) \). At x = −1, |y| = 1, and at x = 1, y = 5.

    \( \displaystyle \frac{1}{2}\times\frac{1}{3}\times 1+\frac{1}{2}\times\frac{5}{3}\times 5=\frac{1}{6}+\frac{25}{6} \)

    \( \displaystyle =\frac{13}{3} \)

Why the other options are incorrect

• Option (1) 4 sq.units — Incorrect. 4 is \( \int_{-1}^{1}(3x+2)\,dx \) taken in one piece; the part below the x-axis then counts as negative instead of adding to the area.

• Option (3) \( \displaystyle\frac{1}{6} \) sq.units — Incorrect. \( \frac{1}{6} \) is only the part below the x-axis, between x = −1 and \( x=-\frac{2}{3} \).

• Option (4) \( \displaystyle\frac{25}{6} \) sq.units — Incorrect. \( \frac{25}{6} \) is only the part above the x-axis, between \( x=-\frac{2}{3} \) and x = 1.

NCERT (2026-27 edition):
Application of Integrals