The area of the region bounded by the line 3y=-2x+7 , x-axis and the…

The correct answer is Option (3) → 25/6

Write the line as \( y=\frac{7-2x}{3} \). It meets the x-axis where \( 7-2x=0 \), that is at \( x=\frac{7}{2} \), which lies between 0 and 4. So the region is in two parts: above the x-axis on \( \left[0,\frac{7}{2}\right] \) and below it on \( \left[\frac{7}{2},4\right] \). The part below must be taken with a positive sign.

    \( \displaystyle A_1=\int_{0}^{7/2}\frac{7-2x}{3}\,dx \)

    \( \displaystyle =\frac{1}{3}\Big[7x-x^2\Big]_{0}^{7/2} \)

    \( \displaystyle =\frac{1}{3}\left(\frac{49}{2}-\frac{49}{4}\right) \)

    \( \displaystyle =\frac{49}{12} \)

    \( \displaystyle A_2=\left|\int_{7/2}^{4}\frac{7-2x}{3}\,dx\right| \)

    \( \displaystyle =\left|\frac{1}{3}\left[(28-16)-\frac{49}{4}\right]\right| \)

    \( \displaystyle =\left|-\frac{1}{12}\right| \)

    \( \displaystyle =\frac{1}{12} \)

    \( \displaystyle \text{Area}=A_1+A_2 \)

    \( \displaystyle =\frac{49}{12}+\frac{1}{12} \)

    \( \displaystyle =\frac{50}{12} \)

    \( \displaystyle =\frac{25}{6} \)

Shortcut: Both parts are triangles. The first has base \( \frac{7}{2} \) and height \( y(0)=\frac{7}{3} \); the second has base \( 4-\frac{7}{2}=\frac{1}{2} \) and height \( |y(4)|=\frac{1}{3} \).

    \( \displaystyle \frac{1}{2}\cdot\frac{7}{2}\cdot\frac{7}{3}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{3}=\frac{49}{12}+\frac{1}{12} \)

    \( \displaystyle =\frac{25}{6} \)

Why the other options are incorrect

• Option (1) 6 — Incorrect. The two parts add up to \( \frac{49}{12}+\frac{1}{12}=\frac{25}{6} \), not 6.

• Option (2) 4 — Incorrect. This is \( \int_{0}^{4}\frac{7-2x}{3}\,dx \) without splitting. The part below the x-axis is negative and cancels part of the area: \( \frac{49}{12}-\frac{1}{12}=4 \).

• Option (4) 7 — Incorrect. The two parts add up to \( \frac{25}{6} \), not 7.