Definite vs Indefinite Integrals

The difference at a glance

• Notation: indefinite $\int f(x)\,dx$; definite $\int_a^b f(x)\,dx$ with a lower limit $a$ and an upper limit $b$.
• Result: indefinite gives a function plus a constant; definite gives a number.
• Meaning: indefinite answers which functions have this derivative; definite answers how much accumulates between $a$ and $b$.
• Constant: indefinite needs $+ C$; in a definite integral the constant cancels.
• Check: differentiate an indefinite answer to recover the integrand; estimate a definite answer from a graph or with rectangles.

If integrals are new to you, start with what is an integral, which builds both ideas from scratch.

Indefinite integrals: families of antiderivatives

The indefinite integral of $f$ is the set of all functions whose derivative is $f$. Because adding a constant does not change a derivative, the answer is always a family of curves, each a vertical shift of the others:

To pick one member of the family you need extra information, usually a known value. This is called an initial value problem.

Physics uses this all the time. If acceleration is constant at $-9.8$ m/s², integrating once gives velocity $v(t) = -9.8t + v_0$, and integrating again gives height $h(t) = -4.9t^2 + v_0 t + h_0$. The two constants are the initial velocity and initial height.

Definite integrals: numbers from limits of sums

A definite integral is defined as a limit of Riemann sums and represents the signed area between the graph and the x-axis from $a$ to $b$. The Fundamental Theorem of Calculus provides the shortcut for evaluating it:

Using $F + C$ instead of $F$ changes nothing, because $(F(b) + C) - (F(a) + C) = F(b) - F(a)$. That is why the constant is dropped.

Properties of definite integrals

Sums and constant multiples pass through the integral sign, just as with derivatives. Symmetry can save a lot of work: if $f$ is odd, meaning $f(-x) = -f(x)$, then $\int_{-a}^{a} f\,dx = 0$. For example, $\int_{-2}^{2} x^3\,dx = 0$ because the negative and positive halves cancel. If $f$ is even, $\int_{-a}^{a} f\,dx = 2\int_0^a f\,dx$, so $\int_{-1}^{1} x^2\,dx = 2\cdot\frac{1}{3} = \frac{2}{3}$.

The average value of $f$ on $[a, b]$ is $\frac{1}{b - a}\int_a^b f\,dx$. For $x^2$ on $[0, 3]$ that is $\frac{1}{3}\cdot 9 = 3$: a rectangle of height 3 and width 3 has the same area as the region under the parabola.

Substitution works differently in each

With an indefinite integral, you substitute, integrate, and then substitute back to the original variable. With a definite integral, you can instead change the limits to match the new variable and never go back.

A common error is substituting $u$ but keeping the old $x$ limits. Either change the limits or convert back to $x$ before evaluating, never half of each.

Where the two meet: accumulation functions

A definite integral with a variable upper limit is a function: $A(x) = \int_a^x f(t)\,dt$. It is a particular antiderivative of $f$, the one that equals 0 at $x = a$. For example, $\int_0^x t^2\,dt = \frac{x^3}{3}$.

The first part of the Fundamental Theorem says $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. So $\frac{d}{dx}\int_1^x \cos(t^2)\,dt = \cos(x^2)$, even though that integral has no elementary formula. If the upper limit is itself a function, the chain rule adds a factor: $\frac{d}{dx}\int_0^{x^2}\sin t\,dt = 2x\sin(x^2)$. Rules for differentiating are summarised in derivative rules explained.

Improper integrals

A definite integral over an infinite interval, or of a function that blows up inside the interval, is called improper and is defined as a limit. For example:

So an infinitely long region can have a finite area. By contrast, $\int_1^{\infty}\frac{1}{x}\,dx = \lim_{b \to \infty}\ln b$ grows without bound, so it diverges. This comparison is the basis of the integral test in series convergence tests.

Which one does a question want?

Look for limits on the integral sign or words such as from, between, total, area or how much. Those signal a definite integral and a numerical answer. Words such as find a function, general solution or antiderivative signal an indefinite integral with $+ C$. A question that gives a starting value, such as $f(0) = 2$, wants an indefinite integral followed by solving for $C$.

Use the integral calculator for indefinite integrals with steps, and the definite integral calculator to evaluate over an interval, including improper integrals.

A worked comparison with one function

Seeing both kinds of integral applied to the same function makes the contrast concrete. Take the function that gives the rate at which water flows into a tank, in litres per minute, and suppose it equals $4t + 2$ after $t$ minutes.

The indefinite integral, $2t^2 + 2t + C$, describes the amount of water in the tank at any time, but only up to an unknown starting amount. If the tank held 10 litres at the start, then $C = 10$ and the volume after $t$ minutes is $2t^2 + 2t + 10$ litres. That is a function you can evaluate at any moment.

The definite integral from $t = 0$ to $t = 5$ answers a different question: how much water flowed in during those five minutes? It equals $(50 + 10) - 0 = 60$ litres, regardless of how much was in the tank to begin with. The first answer is a formula for a state; the second is a single number for a change. Most applied problems fall clearly into one of those two types once you ask whether the question wants a quantity at a time or a total over an interval.

Common mistakes with each type

With indefinite integrals, the most common error is leaving off the constant. It may seem harmless, but in an initial value problem it removes the very thing you are asked to find, and in differential equations it hides an entire family of solutions. A second error is forgetting to rewrite the answer in terms of the original variable after a substitution.

With definite integrals, typical mistakes are subtracting in the wrong order, mixing up old and new limits after a substitution, and reading a signed area as a total area. Another trap is integrating straight across a point where the function is undefined, such as $\frac{1}{x^2}$ from $-1$ to 1. Blindly applying the formula there gives $-2$, which is impossible for a positive function; in fact the integral diverges, because the function blows up at 0. Always check that the integrand is continuous on the whole interval before using the Fundamental Theorem.

How the two ideas fit together

It helps to think of the indefinite integral as the tool and the definite integral as the job. Finding an antiderivative is the hard creative step; once you have one, every definite integral of that function on every interval follows by subtraction. That is why textbooks spend so long on techniques of indefinite integration, and why a single good antiderivative can answer a whole set of area, distance and accumulation questions at once.

Frequently asked questions