91Ó°ÊÓ

To understand why we integrate, we need to grasp how integration helps in finding total areas under curves. The area under a curve from point A to point B gives us crucial values such as probabilities in our context here. For our function: The integral of the function over a given interval tells us the total probability. Hence, we compute: \int_{0}^{1} k x^2 (1-x) \,dx = 1\ Integration handles complex shapes by summing infinitesimally small areas to find the whole. We break the problem down with: \int_{0}^{1} k x^2 (1-x) \,dx = k \int_{0}^{1} (x^2 - x^3) \,dx\ The terms are now easier to work with separately: • \int_{0}^{1} x^2 \,dx = \frac{x^3}{3}≈[0 to 1] = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}\ \( \int_{0}^{1} x^3 \,dx = \frac{x^4}{4}≈\)-=[0 to 1]=\frac{1^4}{4} - \frac{0^1}{4}= \frac{1}{4}\ Combining the 2 terms, we get: \frac{1}{3} - \frac{1}{4}\ which is \frac{4}{12} - \frac{3}{12} = \frac{1}{12}\ This last result is the area under the curve before accounting for 'k'.

A Probability Density Function (PDF) holds key properties that are critical in statistics and probability theory. First and foremost: * The PDF must always integrate over its full range to 1, representing the total probability. * The function must be non-negative for all values in its domain. In our example: \[\int_{0}^{1} k x^2 (1-x) \,dx =1\] illustrates the first property wherein the total \

The normalization constant ensures the PDF integrates to 1. It 'normalizes' our function. \ \ For our problem: \[\int_{0}^{1} f(x) dx=1\] We have solved it up to the point: \[\frac{k}{12}=1\] Solving for k here: \[k = 12\] Without normalization, the total area could be any number but not 1.