Finding a value from a probability density function

Question

I've been asked the following based on a probability course I'm taking :

$$f(x) = \begin{cases} cx^3(1-x)^2 & 0 \le x \le 1 \\ 0 & \text{otherwise.} \end{cases}$$

where $x$ has a beta distribution with probability density function.

Find $c$

My solution :

Step 1 $$ 1 = \int_0^1 cx^3(1-x^2)dx $$

Step 2 $$ 1 = c\int_0^1 x^3(1-x)^2dx $$

Step 3 $$ 1 = c\int_0^1 x^3-x^5 dx $$

As $$ \int x^ndx = \frac{x^{n+1}}{n+1} $$

Therefore :

Step 4 $$ c \Big[ \frac{x^4}{4} - \frac{x^6}{6} \Big]^1_0 $$

My solution is based on my notes but I'm unsure how to arrive at step 4 from step 3. Specifically how the integral symbol is removed and arrive at square brackets with 1 & 0 : $ \Big]^1_0$

How should c be solved from $c\Big[ \frac{x^4}{4} - \frac{x^6}{6} \Big]^1_0 $ ?

Update :

$c\Big[ \frac{x^4}{4} - \frac{x^6}{6} \Big]^1_0 $

$\frac{1^4}{4} - 0 = \frac{1}{4}$

$c = \frac{1}{4}$

Answer 1

You have the right idea that the PDF must integrate to 1, by definition. This gives: $\int_0^1 c x^3 (1-x)^2 dx = 1$, implying $\int_0^1 c x^3(1+x^2-2x) dx = 1$.

Using the general rule $\int_0^1 x^n dx = 1/(n+1)$, you get:

$c(1/4 + 1/6 - 2/5) = 1$, i.e. $ c = 60$.