Continuum random variable distribution: integral and trapezium-rule methods give different result

Question

Suppose p.d.f. is $\frac{1}{63}x^{2}$. Find the $P(4<x<5)$.

I've tried with integration method and trapezium-rule, but they give me different result.

With integration,

$$\int_{4}^{5}\frac{1}{63}x^{2}\:dx=\left.\frac{1}{63}\:\right[_4^5\:\frac{1}{3}x^{3}=\frac{1}{63}\:(\frac{125}{3}-\frac{64}{3})=\frac{61}{189}$$

But with trapezium-rule,

$$(5-4)\cdot\frac{f(5)+f(4)}{2}=1\cdot\frac{(\frac{25}{63}+\frac{16}{63})}{2}=\frac{41}{126}$$

What I missed?

Answer 1

The trapezoidal rule is an approximation. It only gives you the same answer when you partition the interval up into $n$ pieces and take $n\to\infty$ (the limit). Your first solution looks correct.