The area of the region bounded by the line \( y=3x+2 \), the \( x \) -axis and the ordinates \( x=-1 \) and \( x=1 \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{13}{3} \)
The line meets the x-axis where \( 3x+2=0 \), that is, \( x=-\frac{2}{3} \), which lies between −1 and 1. So the region has a part below the x-axis and a part above it, and the two parts must be added as positive areas.
Part below the x-axis, from x = −1 to \( x=-\frac{2}{3} \):
\( \displaystyle A_1=\left|\int_{-1}^{-2/3}(3x+2)\,dx\right| \)
\( \displaystyle =\left|\left[\frac{3x^2}{2}+2x\right]_{-1}^{-2/3}\right| \)
\( \displaystyle =\left|\left(\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{2}-2\right)\right| \)
\( \displaystyle =\left|-\frac{2}{3}+\frac{1}{2}\right| \)
\( \displaystyle =\frac{1}{6} \)
Part above the x-axis, from \( x=-\frac{2}{3} \) to x = 1:
\( \displaystyle A_2=\int_{-2/3}^{1}(3x+2)\,dx \)
\( \displaystyle =\left[\frac{3x^2}{2}+2x\right]_{-2/3}^{1} \)
\( \displaystyle =\frac{7}{2}-\left(-\frac{2}{3}\right) \)
\( \displaystyle =\frac{25}{6} \)
\( \displaystyle \text{Area}=\frac{1}{6}+\frac{25}{6} \)
\( \displaystyle =\frac{26}{6}=\frac{13}{3} \)
Shortcut: Both parts are triangles. At x = −1, y = −1; 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) \( \displaystyle\frac{25}{6} \) — Incorrect. \( \frac{25}{6} \) is only the part above the x-axis; the small triangle below the axis between x = −1 and \( x=-\frac{2}{3} \) is left out.
Option (2) 4 — Incorrect. 4 is \( \int_{-1}^{1}(3x+2)\,dx \) taken in one piece, where the part below the axis is subtracted instead of added.
Option (3) 0 — Incorrect. The region is not empty, so its area cannot be 0.
NCERT (2026-27 edition):
Application of Integrals
