Tomi's Challenges - #19

$\begin{aligned} I & = \int_a^b F'(x) \left(F(x) + \frac 1{2F(x)} \right) dx \\ & = \int_a^b F'(x) F(x)\ dx + \int_a^b \frac {F'(x)}{2F(x)} \ dx \\ & = \frac {F^2(x)}2 + \frac {\ln (F(x))}2 \ \bigg|_a^b \\ & = \frac {F^2(b)-F^2(a)}2 + \frac 12 \ln \left(\frac {F(b)}{F(a)} \right) & \small \blue{\text{Note that }\frac {F(b)}{F(a)}=e^{42}} \\ & = \frac {(F(b)-F(a))(F(b)+F(a))}2 + \frac 12 \cdot 42 & \small \blue{\text{and that }F^2(b)-F^2(a) \ge 0} \\ \implies \min (I) & = \boxed{21} & \small \blue{\text{when }F(b) = F(a)} \end{aligned}$

$\int_{a}^{b} F'(x)\left(F(x)+\frac{1}{2F(x)}\right) \,dx$

$\int_{a}^{b} F'(x)F(x)+\frac{F'(x)}{2F(x)} \,dx$

$\int_{a}^{b} F'(x)F(x)\,dx + \int_{a}^{b} \frac{F'(x)}{2F(x)} \,dx$

Let $I=\int_{a}^{b} F'(x)F(x)\,dx$ . Using integration by parts, we get:

$I=F^2(b)-F^2(a)-\int_{a}^{b} F'(x)F(x)\,dx$

$I=F^2(b)-F^2(a)-I$

By isolating $I$ , we get:

$\int_{a}^{b} F'(x)F(x)\,dx=\frac{F^2(b)-F^2(a)}{2}$

The second integral is solved by recalling that

$\frac{d}{dx}(\log_nF(x))=\frac{F'(x)}{\ln nF(x)}$

Which means that the second integral amounts to just the following:

$\int_{a}^{b} \frac{F'(x)}{2F(x)} \,dx=\log_{e^2}F(b)-\log_{e^2}F(a)=\frac{1}{2}\ln \frac{F(b)}{F(a)}=21$

So the whole integral is equal to:

$\int_{a}^{b} F'(x)\left(F(x)+\frac{1}{2F(x)}\right) \,dx=\frac{F^2(b)-F^2(a)}{2}+21$

But from $\frac{F(b)}{F(a)}=e^{42}$ we have $F(b)=e^{42}F(a)$ , which means that our integral is just the following quadratic function:

$\frac{e^{84}F^2(a)-F^2(a)}{2}+21$

Minimizing this is straightforward, so we will skip it assuming the reader knows how to minimize quadratic functions. This function has a minimum value at $21$ for $F(a)=0$ . However, be careful, $F(a)$ is in the denominator in one of the conditions given in the exercise, so you will need to calculate the minimum value in terms of $F(b)$ , which yields the same results.

I think there is something wrong here. Correct me if I am wrong. I got the 21 part but how can you say that the minimum value of f^2(b)-f^2(a)=0. The explanations, I have seen suggest that f(a) is 0.But ,if f(a) is 0,how come f(b)/f(a) even be finite?