\rm \int_(-\infty)^0√(1+2^2)e^(2\theta)d\theta=\phi-\frac12\\​

Question

`\rm \int_(-\infty)^0√(1+2^2)e^(2\theta)d\theta=\phi-\frac12\\`​

Answer 1

Nothing tricky here:

`\displaystyle \int_(-\infty)^0 √(1+2^2) \, e^(2\theta) \, d\theta = \sqrt5 \int_(-\infty)^0 e^(2\theta) \, d \theta \\\\ ~~~~~~~~ = \frac{\sqrt5}2 \left(e^0 - \lim_(\theta\to-\infty) e^(2\theta)\right) \\\\ ~~~~~~~~ = \frac{\sqrt5}2`

Recall that

`\phi = \frac{1+\sqrt5}2`

and this is exactly 1/2 more than the value of the integral.