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Calculus

What Is an Integral? A Beginner-Friendly Explanation

By MathSolver Pro Editorial Team · Published

An integral adds up infinitely many tiny pieces to find a total: an area, a distance, a volume or any accumulated quantity. The Fundamental Theorem of Calculus links this adding-up process to antiderivatives, which is what makes integrals practical to compute.

Integration as adding up

If a car drives at a constant 60 km per hour for 2 hours, it travels 60×2=12060 \times 2 = 120 km. On a graph of speed against time, that distance is the area of a rectangle of height 60 and width 2. When the speed changes, the region under the graph is no longer a rectangle, but the distance is still the area under the speed curve. Finding that area is what an integral does.

More generally, whenever a quantity is a rate multiplied by an amount, such as speed times time, density times volume or force times distance, and the rate varies, the total is an integral. The derivative breaks a total into rates; the integral rebuilds the total from the rates.

Riemann sums: approximating with rectangles

To find the area under f(x)=x2f(x) = x^2 from x=0x = 0 to x=2x = 2, split the interval into 4 strips of width 0.50.5. Using the right-hand edge of each strip for the height gives rectangles with heights f(0.5)f(0.5), f(1)f(1), f(1.5)f(1.5) and f(2)f(2):

0.5 (0.25+1+2.25+4)=3.750.5\,(0.25 + 1 + 2.25 + 4) = 3.75

Using the left-hand edges instead gives 0.5 (0+0.25+1+2.25)=1.750.5\,(0 + 0.25 + 1 + 2.25) = 1.75. The true area lies between the two. As the number of strips nn grows, both sums approach the same value, and that limit is the definite integral:

∫abf(x) dx=lim⁡n→∞∑i=1nf(xi) Δx\int_a^b f(x)\,dx = \lim_{n \to \infty}\sum_{i=1}^{n} f(x_i)\,\Delta x

The integral sign is a stretched S for sum, f(x)f(x) is the height, and dxdx represents the infinitely thin width. For x2x^2 on [0,2][0, 2] the limit is exactly 83≈2.667\frac{8}{3} \approx 2.667. The idea of a limit, explained in our limits guide, is what turns an approximation into an exact value.

Signed area

Where the graph lies below the x-axis, f(x)f(x) is negative, so the rectangles contribute negative amounts. The definite integral therefore measures signed area: area above the axis minus area below it.

For example, ∫02πsin⁡x dx=0\int_0^{2\pi}\sin x\,dx = 0, because the hump above the axis between 0 and π\pi exactly cancels the one below between π\pi and 2π2\pi. Each hump has area 2, so the total geometric area is 4. To find total area, split the interval where the function changes sign and add the absolute values of the pieces.

Antiderivatives and the indefinite integral

An antiderivative of ff is a function FF whose derivative is ff. Since ddxx3=3x2\frac{d}{dx}x^3 = 3x^2, the function x3x^3 is an antiderivative of 3x23x^2. So is x3+7x^3 + 7, or x3x^3 plus any constant, because constants have derivative zero. The indefinite integral collects them all:

∫3x2 dx=x3+C\int 3x^2\,dx = x^3 + C

Each basic derivative rule, read backwards, gives an integration rule:

∫xn dx=xn+1n+1+C (n≠−1),∫1x dx=ln⁡∣x∣+C,∫ex dx=ex+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\ (n \ne -1), \quad \int \frac{1}{x}\,dx = \ln|x| + C, \quad \int e^x\,dx = e^x + C
∫cos⁡x dx=sin⁡x+C,∫sin⁡x dx=−cos⁡x+C\int \cos x\,dx = \sin x + C, \qquad \int \sin x\,dx = -\cos x + C

Example 1: integrate a polynomial

  1. ∫(3x2−4x+5) dx\int (3x^2 - 4x + 5)\,dx: integrate term by term.
  2. 3⋅x33−4⋅x22+5x+C=x3−2x2+5x+C3\cdot\frac{x^3}{3} - 4\cdot\frac{x^2}{2} + 5x + C = x^3 - 2x^2 + 5x + C.
  3. Check by differentiating: 3x2−4x+53x^2 - 4x + 5. Differentiating your answer is always a reliable check.

The Fundamental Theorem of Calculus

The two meanings of integral, an area found by a limit of sums and a family of antiderivatives, are connected by the Fundamental Theorem of Calculus. Its practical form says that if FF is any antiderivative of a continuous function ff, then:

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

This is remarkable: an infinite sum of rectangles reduces to two function values. The other part of the theorem says that the accumulated area function A(x)=∫axf(t) dtA(x) = \int_a^x f(t)\,dt has derivative f(x)f(x), so integration and differentiation undo each other.

Example 2: two definite integrals

  1. ∫02x2 dx=[x33]02=83−0=83\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}, matching the limit of the Riemann sums.
  2. ∫13(2x+1) dx=[x2+x]13=(9+3)−(1+1)=10\int_1^3 (2x + 1)\,dx = \left[x^2 + x\right]_1^3 = (9 + 3) - (1 + 1) = 10.
  3. The constant CC cancels in F(b)−F(a)F(b) - F(a), so it can be left out of definite integrals.

The difference between the two kinds of integral is explored further in definite vs indefinite integrals.

Two techniques beyond the basic rules

Integration has no product or chain rule that works mechanically every time, so some techniques reverse those derivative rules.

Example 3: substitution, the chain rule in reverse

  1. ∫2x(x2+1)3 dx\int 2x(x^2 + 1)^3\,dx: let u=x2+1u = x^2 + 1, so du=2x dxdu = 2x\,dx.
  2. The integral becomes ∫u3 du=u44+C\int u^3\,du = \frac{u^4}{4} + C.
  3. Substitute back: (x2+1)44+C\frac{(x^2 + 1)^4}{4} + C.

Example 4: integration by parts, the product rule in reverse

  1. The formula is ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du.
  2. For ∫xex dx\int xe^x\,dx, choose u=xu = x and dv=ex dxdv = e^x\,dx, so du=dxdu = dx and v=exv = e^x.
  3. ∫xex dx=xex−∫ex dx=ex(x−1)+C\int xe^x\,dx = xe^x - \int e^x\,dx = e^x(x - 1) + C.

Applications

  • Distance from velocity: if v(t)=3t2v(t) = 3t^2 m/s, the displacement over the first 2 seconds is ∫023t2 dt=8\int_0^2 3t^2\,dt = 8 m.
  • Area between curves: between y=xy = x and y=x2y = x^2 on [0,1][0, 1] the area is ∫01(x−x2) dx=12−13=16\int_0^1 (x - x^2)\,dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.
  • Average value: the average of ff on [a,b][a, b] is 1b−a∫abf(x) dx\frac{1}{b - a}\int_a^b f(x)\,dx.
  • Probability: areas under a probability density curve give probabilities, as with the normal curve behind z-scores.
  • Volumes, work done by a variable force, total cost from marginal cost and centres of mass are all integrals.

Common mistakes

  • Forgetting +C+ C on an indefinite integral.
  • Applying the power rule to 1x\frac{1}{x}, which would divide by zero; its integral is ln⁡∣x∣\ln|x|.
  • Thinking ∫fg dx\int f g\,dx equals ∫f dx⋅∫g dx\int f\,dx \cdot \int g\,dx.
  • Reading a definite integral as total area when part of the graph is below the axis.
  • Computing F(a)−F(b)F(a) - F(b) instead of F(b)−F(a)F(b) - F(a).

The integral calculator finds antiderivatives with each step shown, and the definite integral calculator evaluates integrals over an interval and reports the numerical value.

Why the Fundamental Theorem makes sense

Return to the car. Suppose s(t)s(t) is its position and v(t)=s′(t)v(t) = s'(t) its velocity. Adding up velocity times tiny time intervals over a journey gives the total change in position, which is the final position minus the starting position. In symbols, ∫abv(t) dt=s(b)−s(a)\int_a^b v(t)\,dt = s(b) - s(a), which is exactly the theorem with ss playing the role of the antiderivative.

Seen this way, the theorem is almost common sense: adding up all the small changes in a quantity gives its total change. What makes it powerful is that it turns hard limit calculations into easy subtraction, as long as an antiderivative can be found.

Numerical integration

When no antiderivative can be written down, or when the function is only known from measurements, integrals are estimated numerically. Better estimates come from better shapes than rectangles. The trapezoidal rule averages the left and right sums; for x2x^2 on [0,2][0, 2] with 4 strips it gives 1.75+3.752=2.75\frac{1.75 + 3.75}{2} = 2.75. The midpoint rule uses the height at the centre of each strip and gives 0.5 (0.0625+0.5625+1.5625+3.0625)=2.6250.5\,(0.0625 + 0.5625 + 1.5625 + 3.0625) = 2.625. Both are much closer to the exact 2.6672.667 than either one-sided sum.

Simpson's rule, which fits parabolas through groups of three points, is exact for this function and for any cubic. Software uses refinements of these ideas to compute integrals to many decimal places in a fraction of a second.

A short history

Finding areas by adding thin slices is ancient. Archimedes found the area of a parabolic segment more than two thousand years ago using a method close to a limit. The breakthrough of the seventeenth century, credited to Newton and Leibniz, was the Fundamental Theorem, which showed that areas could be found by reversing differentiation. Bernhard Riemann gave the sums their modern precise definition in the nineteenth century, which is why they carry his name.

Frequently asked questions

What is the difference between a derivative and an integral?

A derivative gives a rate of change at an instant; an integral adds up rates to give a total. The Fundamental Theorem of Calculus shows they are inverse processes, in the same way multiplication and division are.

Why does an indefinite integral have + C?

Every constant has derivative zero, so if FF is an antiderivative of ff, so is F+CF + C for any constant CC. The +C+ C represents that whole family.

Why can an integral come out as a negative number?

Yes. It measures signed area, so regions below the x-axis count as negative. A negative result means more area lies below the axis than above it on that interval.

Does every function have an antiderivative I can write down?

No. Every continuous function has an antiderivative, but some, such as that of e−x2e^{-x^2}, cannot be written with elementary functions. Their definite integrals are computed numerically.

What does the dx at the end of an integral mean?

It names the variable you are integrating with respect to and stands for the width of the thin strips in the Riemann sum. It also matters in substitution, where dxdx is converted to dudu, and in integrals with several letters, where it tells you which one is the variable.

MathSolver Pro Editorial Team

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Our editorial team writes every guide to explain the reasoning behind each method, not just the answer. Each worked example is checked against the site's own solver engines and automated tests before it is published, and articles are revised whenever a method or an exam format changes.