Continuous Distributions: Areas Under the PDF

Discrete vs. continuous

For a continuous random variable, individual values have probability zero (there are uncountably many). Instead, we specify a probability density function (PDF) $f(x)$, and probabilities are areas under the curve: $P(a \le X \le b) = \int_a^b f(x) \, dx$

Any valid PDF satisfies $f(x) \ge 0$ and $\int_{-\infty}^{\infty} f(x) \, dx = 1$.

Because $P(X = a) = \int_a^a f \, dx = 0$, strict and non-strict inequalities give the same probability: $P(a < X < b) = P(a \le X \le b)$.

Step-by-step solution

The PDF: $f(x) = 2x$ on $[0, 1]$, and $0$ elsewhere.

Step 1 — Verify it is a valid PDF. $\int_0^1 2x \, dx = \left[ x^2 \right]_0^1 = 1 \checkmark$

and $f(x) = 2x \ge 0$ on $[0, 1]$. ✓

Step 2 — Set up the target probability. $P(0.3 < X < 0.7) = \int_{0.3}^{0.7} 2x \, dx$

Step 3 — Evaluate the integral. $\int 2x \, dx = x^2 + C$ $\left[ x^2 \right]_{0.3}^{0.7} = 0.49 - 0.09 = \boxed{0.40}$

So there is a 40% chance that $X$ falls between $0.3$ and $0.7$.

Sanity check

The PDF grows linearly, so more mass sits near $x = 1$. Our interval $[0.3, 0.7]$ contains the middle — enough of the bulk for $\approx 40\%$ to feel right. Compare: $P(0 < X < 0.5) = 0.25$ (less, because the left tail is low density) and $P(0.5 < X < 1) = 0.75$ (more, because the right tail is denser). All consistent.

The cumulative distribution function (CDF)

$F(x) = P(X \le x) = \int_{-\infty}^x f(t) \, dt$. For our PDF: $F(x) = \begin{cases} 0 & x < 0 \\ x^2 & 0 \le x \le 1 \\ 1 & x > 1 \end{cases}$

Then $P(0.3 < X < 0.7) = F(0.7) - F(0.3) = 0.49 - 0.09 = 0.40$, matching the integral.

Key moments

Expected value: $E(X) = \int_0^1 x \cdot 2x \, dx = \int_0^1 2x^2 \, dx = \tfrac{2}{3}$

Variance: $E(X^2) - (E(X))^2 = \int_0^1 2x^3 \, dx - (2/3)^2 = \tfrac{1}{2} - \tfrac{4}{9} = \tfrac{1}{18}$

Common mistakes

• Treating $f(x)$ like a probability. It is a density — values can exceed 1 (e.g. $f(0.9) = 1.8$ here). Only the area is a probability.
• Forgetting the endpoint doesn't matter. For continuous $X$, $P(X = 0.3) = 0$, so strict vs. non-strict inequality is identical.
• Integrating over the wrong interval. If $f$ is zero outside $[0, 1]$, an integral that strays outside just contributes zero — but be explicit about the support.

Try it in the visualization

Drag the two endpoints $a$ and $b$ along the PDF curve. The shaded area shades in real time, and a readout shows the corresponding probability computed from $F(b) - F(a)$. Toggle to a different PDF (uniform, linear, or triangular) and watch how the shape of $f$ changes the answer.