Finding constant in Probability density function.

sid9221
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A continuous random variable X has pdf:

f_X(x)=\left \{ k(x+3), 0\leq x\leq 1\right \}

0 otherwise.

Find k.

I solved the integral (from 0..1) and solved for the result equal to 1.

Hence I got k=2/7.

Is this the right way to proceed as the question continues and I want to check if I'm starting correctly
 
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yep, you just want to normalise to 1
 
What do you mean normalize to 1 ?

It was 7/2k=1
So k=2/7

?
 
sid9221 said:
What do you mean normalize to 1 ?

It was 7/2k=1
So k=2/7

?

Normalizing means making \int_0^1 f_X(x) \, dx = 1.

RGV
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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