Find the C of an integrated function

Question

The function $F : [1, 4]\to\mathbb R$ is given by $$F(\theta) = \int_1^4\left|\ln^2(x) - \ln^2(\theta)\right|\frac{\mathrm dx}x, 1\le\theta\le4.$$

There is a $C$ such that $$F(\theta) = \frac43\ln^3(\theta) - 2\ln(2)\ln^2(\theta) + C, \forall\theta\in[1, 4].$$ [You don't have to prove this.] Find $C$.

We are given an integral which is already solved for us. Then we have to solve for a $C$ so that theta is an element of $[1,4]$.

I need help understanding the question, I just can't seem to understand what the exact conditions are.

Any help is appreciated, thank you.

Answer 1

Assuming you know how to integrate the expression, we have

$$F(\theta) = \int_1^4 \left|\ln^2(x) - \ln^2(\theta)\right|\frac{\mathrm dx}x = \frac13\ln^3(4) - \ln(4)\ln^2(\theta)$$

So,

$$\frac13\ln^3(4) - \ln(4)\ln^2(\theta) = F(\theta) = \frac43\ln^3(\theta) - 2\ln(2)\ln^2(\theta) + C$$

Set $\theta = 1 \implies \ln^n(1) = 0$ to get

$$\frac13\ln^3(4) - \ln(4)\cdot 0 = F(1) = \frac43\cdot 0 - 2\ln(2)\cdot 0 + C$$

Finally,

$$C = \frac13\ln^3(4)$$

Answer 2

You know that:

There is a $C$ such that $$F(\theta) = \frac43\ln^3(\theta) - 2\ln(2)\ln^2(\theta) + C, \forall\theta\in[1, 4].$$

By calculating F(1):

$F(1)=C$

However

$F(1)=\int_1^4 |\ln^2(x)-\ln^2(1)|\frac{1}{x} dx=\int_1^4 \frac{\ln^2(x)}{x} dx$

As $x\mapsto \frac1x$ is the derivative of $\ln$, the integral is of the form $\int u^2 u'$

Then $F(1)=\left[\frac{\ln^3(x)}{3}\right]_1^4=\frac{\ln^3(4)}{3}$

Hence $C=\frac{\ln^3(4)}{3}$