The area (in sq.units) bounded by the lines \( y=2x+3 \) , \( y=0 \) between \( x=-2 \) and \( x=1 \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{13}{2} \)
The line \( y=2x+3 \) meets the x-axis at \( x=-\frac{3}{2} \), which lies between −2 and 1. The line is below the axis on \( [-2,-\frac{3}{2}] \) and above it on \( [-\frac{3}{2},1] \), so the two parts are taken separately:
\( \displaystyle A_1=\left|\int_{-2}^{-3/2}(2x+3)\,dx\right| \)
\( \displaystyle =\left|\Big[x^2+3x\Big]_{-2}^{-3/2}\right| \)
\( \displaystyle =\left|-\frac{9}{4}-(-2)\right| \)
\( \displaystyle =\frac{1}{4} \)
\( \displaystyle A_2=\int_{-3/2}^{1}(2x+3)\,dx \)
\( \displaystyle =\Big[x^2+3x\Big]_{-3/2}^{1} \)
\( \displaystyle =4-\left(-\frac{9}{4}\right) \)
\( \displaystyle =\frac{25}{4} \)
\( \displaystyle \text{Area}=\frac{1}{4}+\frac{25}{4} \)
\( \displaystyle =\frac{13}{2} \)
Shortcut: Both parts are triangles. On the left the base is \( \frac{1}{2} \) and the height is |y(−2)| = 1, area \( \frac{1}{4} \); on the right the base is \( \frac{5}{2} \) and the height is y(1) = 5, area \( \frac{25}{4} \). Total \( \frac{13}{2} \).
Tip: Integrating straight from −2 to 1 gives 6, because the part below the axis is subtracted. Always check where the line crosses the x-axis before finding an area.