# Find the C of an integrated function

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.

• You can obtain C, by calculating $F(1)=\int_1^4 \frac{\ln^2(x)}{x}dx$ Jan 23, 2020 at 9:29
• @Jean-ClaudeColette what's the logic behind this? Jan 23, 2020 at 9:38
• $F(1)=\frac43 \ln^3(1)-2\ln(2)\ln^2(1)+C=C$. So you will obtain C. Jan 23, 2020 at 9:49
• Appreciate it, thanks. Jan 23, 2020 at 9:52

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,

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

• That explains a lot, thank you! Jan 23, 2020 at 9:48

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

• Thank you once again. Jan 23, 2020 at 10:07