If the area above x-axis, bounded by the curves $y = 3^{βx}, x = 0$ and $x = 3$ is $\frac{26}{\log_e3}$ then the value $β$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $β=1$
Given$\text{Area}=\int_{0}^{3}3^{\beta x}\,dx=\frac{26}{\ln 3}$
$\int_{0}^{3}3^{\beta x}\,dx=\int_{0}^{3}e^{\beta x\ln 3}\,dx=\frac{1}{\beta\ln 3}\big(3^{3\beta}-1\big)$
$\frac{3^{3\beta}-1}{\beta\ln 3}=\frac{26}{\ln 3}$
$\frac{3^{3\beta}-1}{\beta}=26$
$3^{3\beta}-1=26\beta$
Check $\beta=1$: $3^{3}-1=27-1=26=26\cdot 1$
Answer$\beta=1$