The area (in sq. units) of the region bounded by the line \( 2y=3x+6 \), x-axis and the ordinates \( x=-3 \) and \( x=1 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{15}{2} \)
The line is \( y=\frac{3x+6}{2} \). It meets the x-axis at x = −2, so the region has a part below the axis (from x = −3 to x = −2) and a part above it (from x = −2 to x = 1). Each part is taken with a positive sign:
\( \displaystyle \text{Area}=\left|\int_{-3}^{-2}\frac{3x+6}{2}\,dx\right|+\int_{-2}^{1}\frac{3x+6}{2}\,dx \)
\( \displaystyle \int\frac{3x+6}{2}\,dx=\frac{3x^2}{4}+3x \)
\( \displaystyle \int_{-3}^{-2}\frac{3x+6}{2}\,dx=(3-6)-\left(\frac{27}{4}-9\right) \)
\( \displaystyle =-3+\frac{9}{4} \)
\( \displaystyle =-\frac{3}{4} \)
\( \displaystyle \int_{-2}^{1}\frac{3x+6}{2}\,dx=\left(\frac{3}{4}+3\right)-(3-6) \)
\( \displaystyle =\frac{15}{4}+3 \)
\( \displaystyle =\frac{27}{4} \)
\( \displaystyle \text{Area}=\frac{3}{4}+\frac{27}{4} \)
\( \displaystyle =\frac{15}{2} \)
Shortcut: Both parts are right triangles with a vertex at (−2, 0). At x = −3, y = −3/2; at x = 1, y = 9/2.
\( \displaystyle \text{Area}=\frac{1}{2}\times 1\times\frac{3}{2}+\frac{1}{2}\times 3\times\frac{9}{2} \)
\( \displaystyle =\frac{3}{4}+\frac{27}{4}=\frac{15}{2} \)
Tip: If the curve crosses the x-axis inside the interval, split the integral there. A single integral from −3 to 1 would give 27/4 − 3/4 = 6, because the part below the axis cancels part of the area.