The area bounded by \(y = 3x + 1\), \(x = 0\), \(y = 0\) and \(x = a\) is 8 Sq. units. Then value of \(a\) (where \(a \gt 0\)) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
For \(0 \le x \le a\), the line \(y = 3x + 1\) lies above the x-axis, so the area is the definite integral
\( \displaystyle \int_{0}^{a} (3x + 1)\,dx = \left[\frac{3x^2}{2} + x\right]_{0}^{a} = \frac{3a^2}{2} + a \)
Setting this equal to 8:
\(3a^2 + 2a - 16 = 0\)
\((3a + 8)(a - 2) = 0\)
\( \displaystyle a = 2 \text{ or } a = -\frac{8}{3} \)
Since \(a \gt 0\), \(a = 2\).
Shortcut: The region is a trapezium with parallel sides 1 (at \(x = 0\)) and \(3a + 1\) (at \(x = a\)) and width a. For \(a = 2\) its area is \(\frac{1}{2}(1 + 7) \times 2 = 8\).