If the area above x-axis, bounded by the curves \(y = 3^{\beta x}, x = 0\) and \(x = 3\) is \(\displaystyle\frac{26}{\log_e 3}\), then the value \(\beta\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\beta = 1\)
Since \(3^{\beta x} \gt 0\), the curve lies above the x-axis, and the area between x = 0 and x = 3 is the definite integral. Using \(\int a^{kx}\,dx = \frac{a^{kx}}{k \log_e a}\):
\( \displaystyle \text{Area} = \int_{0}^{3} 3^{\beta x}\,dx = \left[\frac{3^{\beta x}}{\beta \log_e 3}\right]_{0}^{3} \)
\( \displaystyle = \frac{3^{3\beta} - 1}{\beta \log_e 3} = \frac{27^{\beta} - 1}{\beta \log_e 3} \)
Equating this to the given area:
\( \displaystyle \frac{27^{\beta} - 1}{\beta} = 26 \)
Putting in the values offered:
\(\beta = 1:\quad 27 - 1 = 26\) (satisfied)
\( \displaystyle \beta = 2:\quad \frac{729 - 1}{2} = 364 \)
\( \displaystyle \beta = -1:\quad \frac{\frac{1}{27} - 1}{-1} = \frac{26}{27} \)
\( \displaystyle \beta = \frac{1}{2}:\quad \frac{\sqrt{27} - 1}{\frac{1}{2}} = 2(3\sqrt{3} - 1) \approx 8.39 \)
So \(\beta = 1\).