Definite Integral

A definite integral is a number that represents the net signed area between the graph of a function \(f(x)\) and the \(x\)-axis, over a fixed interval from \(x = a\) to \(x = b\). Unlike an indefinite integral, which produces a family of antiderivatives plus a constant \(C\), a definite integral always evaluates to a single, specific value.

The definite integral of \(f(x)\) from \(a\) to \(b\) is written as

\(\int_a^b f(x)\,dx\)

Here \(a\) is the lower limit of integration and \(b\) is the upper limit. Any area below the \(x\)-axis counts as negative, so the definite integral gives the net area, not simply the total area between the curve and the axis.

The definite integral is formally defined as a limit of sums. Slice the interval \([a, b]\) into \(n\) thin rectangles, add up their areas, and let \(n\) approach infinity, this is exactly the idea behind a Riemann sum. As the rectangles get infinitely thin, the sum of their areas approaches the exact value of \(\int_a^b f(x)\,dx\). This limiting process is why the definite integral gives an exact area rather than an approximation.

Computing a definite integral directly from a Riemann sum is slow, so in practice you use the fundamental theorem of calculus. If \(F(x)\) is an antiderivative of \(f(x)\), then

\(\int_a^b f(x)\,dx = F(b) - F(a)\)

This means evaluating a definite integral is really a two-step process: find an antiderivative of the function (review antiderivatives if you need a refresher), then plug in the upper and lower limits and subtract.

These rules make evaluating and manipulating definite integrals much easier:

• Reversing the limits flips the sign: \(\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx\)
• A zero-width interval gives zero area: \(\int_a^a f(x)\,dx = 0\)
• Additivity over intervals: \(\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx\)
• Linearity: \(\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx\)
• Constant multiple rule: \(\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx\)

Evaluate \(\int_0^2 x^2\,dx\).

Step 1: Find an antiderivative of \(x^2\), which is \(\frac{x^3}{3}\).

Step 2: Apply the fundamental theorem of calculus.

\(\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}\)

So the shaded region shown above has an area of exactly \(\frac{8}{3}\) square units.

A related idea is what happens when you differentiate a definite integral whose upper limit is a variable, such as \(\frac{d}{dx}\int_a^x f(t)\,dt\). By the fundamental theorem of calculus, this simply equals \(f(x)\). If the upper limit is itself a function \(g(x)\), you multiply by \(g'(x)\) using the chain rule. This connection between differentiation and integration is one of the most useful shortcuts in calculus.

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