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The area of the region bounded by the line $y = 3x + 2$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ is
A$4$
B$\frac{25}{6}$
C$\frac{1}{6}$
D$\frac{13}{3}$
Answer & Solution
Correct answer: D. $\frac{13}{3}$
1. The line meets the $x$-axis at $x = -\frac{2}{3}$; it lies below the axis on $\left(-1, -\frac{2}{3}\right)$ and above on $\left(-\frac{2}{3}, 1\right)$.
2. So area $= \left|\int_{-1}^{-2/3}(3x+2)\,dx\right| + \int_{-2/3}^{1}(3x+2)\,dx$ with antiderivative $\frac{3x^2}{2} + 2x$.
3. The first (below) piece gives $\frac{1}{6}$ in magnitude; the second (above) piece gives $\frac{25}{6}$.
4. Adding, $\frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}$.
5. Forgetting the absolute value and subtracting would lose the sub-axis part, so $\frac{25}{6}$ alone is wrong.
6. Ignoring the split entirely (plain $\int_{-1}^{1}$) gives $4$, which under-counts the area below the axis.
_Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.5_
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