A continuous random variable's probability lives entirely in area under a curve. That curve's domain is often the whole real line. Both facts mean this course leans on the definite integral constantly, and just as often on its improper cousin: an integral with an infinite bound.
An improper integral asks a different question than an ordinary definite integral. Instead of the area over one fixed strip, it asks what the area settles toward as the right edge slides out forever. Some curves fall off fast enough that the strip's area keeps adding smaller and smaller slivers. That total settles on a fixed number, so the integral converges. Others don't fall off fast enough, and the area just keeps growing without any ceiling; that integral diverges instead. The only way to tell which is happening is to compute the limit, not to eyeball the picture.
The definite integral \int_a^b f(x)\,dx is the signed area between f and the x-axis over [a,b]. A definite integral over a finite interval always exists for any continuous function, so no convergence question ever arises there. An improper integral replaces a finite bound with \infty (or -\infty) and is defined as a limit: \int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx. Pushing one bound out to infinity removes that automatic guarantee: the running total might keep growing forever instead of settling down, so existence has to be checked, not assumed. If the limit exists and is finite, the improper integral converges to it; otherwise it diverges, and no value is assigned. The same convergence-or-divergence question also arises with a lower bound of -\infty, or both bounds infinite at once, handled the same way: split the integral and take each limit separately.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
An improper integral \int_a^\infty f(x)\,dx is defined as:
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"f(\infty)" isn't a meaningful expression -- infinity isn't a number you evaluate a function at.
Plugging \infty directly into an antiderivative is exactly the illegal operation this definition avoids.
It has a precise, rigorous definition -- as a limit of ordinary definite integrals.
Why
The infinite bound is always handled by replacing it with a finite variable and taking a limit, never by direct substitution.
An infinite bound is never plugged in directly -- what replaces it before a limit is taken?
From Calculus II — Integral Calculus & Infinite Series
Now the mirror image -- what does \displaystyle\int_{-\infty}^{0} e^{x}\,dx do?
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The limit does exist and is finite here -- this integral is convergent.
The limit settles at a positive value, not zero.
A negative value isn't possible here, since e^x>0 everywhere.
Why
\lim_{t\to-\infty}\int_t^0 e^x\,dx = \lim_{t\to-\infty}\left(1-e^t\right) = 1-0 = 1.
Replace -\infty with a finite t, find an antiderivative of e^x, evaluate at 0 and t, then take the limit as t\to-\infty.
Push the bound toward infinity
Drag the slider to send t further out and watch whether the shaded area settles on a number or keeps climbing.
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A definite integral, built from rectangles
Slide n up and watch the rectangle approximation converge to the exact area under the curve.
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Worked example
A convergent improper integral
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1
Evaluate \displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx.
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2
Rewrite as a limit and find the antiderivative first: \lim_{t\to\infty}\left[-\frac1x\right]_1^t = \lim_{t\to\infty}\left(-\frac1t + 1\right).
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3
As t \to \infty, -\frac1t \to 0, leaving
\boxed{\int_1^{\infty}\frac{1}{x^2}\,dx = 1}
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Worked example
A divergent improper integral
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1
Evaluate \displaystyle\int_1^{\infty} \frac{1}{x}\,dx.
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2
\lim_{t\to\infty}\big[\ln x\big]_1^t = \lim_{t\to\infty}\big(\ln t - 0\big), and \ln t grows without bound as t \to \infty.
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3\boxed{\text{Diverges — the area has no finite total.}}
Where this goes next
Chapter 7 (Continuous random variables) and Chapter 8 (Named continuous distributions)
Every continuous pdf on an unbounded domain, including the exponential and normal distributions, is valid only because its total area works out to exactly 1. Checking that is literally evaluating an improper integral. The exponential distribution's own defining property is built from exactly the convergent-integral pattern practiced here.
Check your understanding (counts toward your progress)
A density-like function satisfies \displaystyle\int_0^{\infty} f(x)\,dx = 1 and \displaystyle\int_0^{3} f(x)\,dx = 0.6. What is \displaystyle\int_3^{\infty} f(x)\,dx?
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Why
The total area splits at x=3: \int_0^3 f + \int_3^\infty f = \int_0^\infty f, so \int_3^\infty f = 1-0.6=0.4. (This is exactly how a later chapter computes tail probabilities.)
Correct answer
Also accepted:
Why
Use the same additive property as a finite interval — split the infinite integral at x=3 and subtract.
A student evaluates \displaystyle\int_{6}^{23} (2x)\,dx as F(6)-F(23) instead of F(23)-F(6). What's the effect?
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Why
F(a)-F(b) = -(F(b)-F(a)) — reversing bound order in the subtraction negates the entire answer.
A student evaluates \displaystyle\int_{6}^{21} (2x)\,dx as F(6)-F(21) instead of F(21)-F(6). What's the effect?
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Why
F(a)-F(b) = -(F(b)-F(a)) — reversing bound order in the subtraction negates the entire answer.
Evaluate \displaystyle\int_0^2 3\,dx (the area of a rectangle of height 3 over width 2).
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Why
The definite integral of a constant c from a to b is c(b-a): 3(2-0) = 6.
Correct answer
Also accepted:
Why
Check your understanding (counts toward your progress)
Evaluate \displaystyle\int_1^\infty \dfrac{1}{x^{23}}\,dx.
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Why
\lim_{t\to\infty}\left[\dfrac{x^{-22}}{-22}\right]_1^t = \dfrac{1}{22} \approx 0.0455 (the t^{-22} term vanishes as t\to\infty).
Correct answer
Also accepted:
Why
Antidifferentiate, then take the limit as the upper bound goes to infinity.
Evaluate \displaystyle\int_1^\infty \dfrac{1}{x^{49}}\,dx.
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Why
\lim_{t\to\infty}\left[\dfrac{x^{-48}}{-48}\right]_1^t = \dfrac{1}{48} \approx 0.0208 (the t^{-48} term vanishes as t\to\infty).
Correct answer
Also accepted:
Why
Antidifferentiate, then take the limit as the upper bound goes to infinity.
Evaluate \displaystyle\int_1^{\infty} \frac{1}{x^3}\,dx.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
The antiderivative is -1/(2x^2); evaluated as a limit, \lim_{t\to\infty}\left(-\frac{1}{2t^2}+\frac{1}{2}\right) = \frac12.
Correct answer
Also accepted:
Why
Replace the infinite bound with t, integrate normally, then take the limit as t → ∞.
Explain what it means for an improper integral to 'converge.'
AI grading isn't configured on this deployment.
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First mistake
Correct answer
Also accepted:
Why
Why
It means the limit defining the integral (as the bound is pushed toward infinity, or toward a discontinuity) exists and equals some finite number — the area under the curve settles on a fixed total instead of growing without bound.
Practice: Evaluating definite integrals (4 problems · doesn’t affect your progress)
Which best describes what \displaystyle\int_a^b f(x)\,dx represents when f(x)\ge 0 on [a,b]?
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That's just f(b), one function VALUE — a definite integral accumulates area over a whole interval, not a single point.
Slope is the DERIVATIVE's job — the integral measures accumulated area, not a rate of change.
This isn't about counting x-values at all — it's a geometric area (or, more generally, an accumulated total).
Why
For a nonnegative f, the definite integral over [a,b] is exactly the area trapped between the curve and the x-axis over that interval.
The area between the curve y=f(x) and the x-axis, from x=a to x=b
Evaluate \displaystyle\int_0^4 5\,dx.
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Why
The integral of a constant c from a to b is c(b-a): 5(4-0)=20.
Correct answer
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Why
Find the value of \displaystyle\int_0^2 3x^2\,dx.
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Why
Antiderivative is x^3; evaluate: 2^3-0^3=8.
Correct answer
Also accepted:
Why
Find the antiderivative of 3x^2 first (it's x^3), then plug in the limits.
A function satisfies \displaystyle\int_0^{\infty} g(x)\,dx = 1. If \displaystyle\int_0^{1} g(x)\,dx = 0.2 and \displaystyle\int_1^{2} g(x)\,dx = 0.3, what is \displaystyle\int_2^{\infty} g(x)\,dx?
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Why
The three pieces (0 to 1, 1 to 2, and 2 to ∞) split the total: 0.2+0.3+\int_2^\infty g = 1, so \int_2^\infty g = 1-0.5=0.5.
Correct answer
Also accepted:
Why
Add up the two known finite pieces first, then subtract that total from the whole (1).
Practice: Convergent versus divergent improper integrals (4 problems · doesn’t affect your progress)
What makes an integral 'improper' rather than an ordinary definite integral?
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A negative value is completely normal for an ordinary definite integral — it isn't what makes one 'improper.'
Plenty of ordinary definite integrals have polynomial integrands; that's unrelated to improperness.
Improper integrals absolutely can be evaluated — that's exactly what taking the limit accomplishes.
Why
An improper integral replaces a finite bound with infinity (or has an integrand that's undefined somewhere in the interval), which is why it needs a limit to define its value at all.
One or both bounds is infinite (or the integrand blows up in the interval)
Find the value of \displaystyle\int_1^{\infty}\frac{1}{x^3}\,dx.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Antiderivative -\frac{1}{2x^2}; as a limit, \lim_{t\to\infty}\left(-\frac1{2t^2}+\frac12\right)=\frac12.
Correct answer
Also accepted:
Why
Replace the infinite bound with t, integrate normally, then take the limit as t goes to infinity.
Compute \displaystyle\int_1^{\infty}\frac{1}{x^4}\,dx.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Antiderivative -\frac{1}{3x^3}; as a limit, \lim_{t\to\infty}\left(-\frac1{3t^3}+\frac13\right)=\frac13.
Correct answer
Also accepted:
Why
Find the antiderivative, then take the limit as the upper bound goes to infinity.
Which of these improper integrals fails to converge?
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This one converges — exponent 2 shrinks the integrand fast enough.
This one converges too — exponent 3 shrinks it even faster.
e^{-x} decays fast enough to converge to a finite value.
Why
\int_1^{\infty} x^{-1/2}\,dx = \lim_{t\to\infty}\left[2\sqrt{x}\right]_1^t, and 2\sqrt{t} grows without bound as t\to\infty — this is the one that diverges (the exponent 1/2 doesn't shrink the integrand fast enough).
Find the antiderivative of each and check whether it settles on a finite value or keeps growing as the bound goes to infinity.
\int_1^{\infty}\frac{1}{\sqrt{x}}\,dx