Integrals
Add up thin slices to find areas and totals.
Derivatives break a curve into tiny slopes. Integrals do the opposite: they add up tiny pieces to get a whole. Slice the region under a curve into thin rectangles, add their areas, and let the slices get thinner. The limit of those sums is the definite integral.
On this page: the area problem, Riemann sums with an interactive explorer, definite integrals and signed area, properties, antiderivatives and the basic rules, u-substitution, a first look at the Fundamental Theorem, five worked examples, nine practice problems and an eight-question quiz.
Section IThe area problem
The area of a rectangle is width times height. But what is the area of the region under y = x², above the x-axis, from x = 0 to x = 2? Its top edge is curved, so no school formula fits.
The idea is to approximate. Cut the interval into n strips of width Δx, stand a rectangle on each strip, and add up the rectangles. With 4 strips of width 0.5 and heights taken at the right end of each strip, the total is 0.5(0.25 + 1 + 2.25 + 4) = 3.75. That overshoots, because each rectangle pokes above the curve. With more, thinner strips the error shrinks, and the totals close in on one number: the exact area, 8/3.
The same adding-up works for anything that builds up from small pieces: distance from speed, water from a flow rate, work from a force.
Section IIRiemann sums
A Riemann sum for f on a ≤ x ≤ b splits the interval into n pieces of width Δx = (b − a)/n, picks a sample point xᵢ* in each piece, and adds the products:
Σ f(xᵢ*) Δx = f(x₁*)Δx + f(x₂*)Δx + … + f(xₙ*)ΔxThe sample point can be the left end, the right end or the midpoint of each piece. The trapezoid rule instead joins the two ends of each piece with a straight line, which is the same as averaging the left and right sums.
Try the interactive version at jugami.com/calculus/integrals/
Things to notice: for x², which is increasing, left sums are always too small and right sums too big. Midpoint sums are usually much closer than either end. Every method closes in on the exact value as n grows.
Section IIIThe definite integral and signed area
The definite integral of f from a to b is the limit of the Riemann sums as the pieces get thinner:
∫ₐᵇ f(x) dx = limn→∞ Σ f(xᵢ*) ΔxThe tall ∫ is a stretched S, for “sum.” The numbers a and b are the limits of integration, f(x) is the integrand, and dx records the variable and stands for the thin width Δx in the limit. Every continuous function on a ≤ x ≤ b has a definite integral.
When f dips below the x-axis, f(xᵢ*)Δx is negative, so those pieces subtract. The definite integral is a signed area: area above the axis minus area below it.
Section IVProperties of definite integrals
- Zero width: ∫ₐᵃ f(x) dx = 0.
- Reversing the limits: ∫ba f(x) dx = −∫ₐᵇ f(x) dx.
- Constant multiple: ∫ₐᵇ k f(x) dx = k ∫ₐᵇ f(x) dx.
- Sum: ∫ₐᵇ (f + g) dx = ∫ₐᵇ f dx + ∫ₐᵇ g dx.
- Joining intervals: ∫ₐc f dx + ∫cb f dx = ∫ₐᵇ f dx.
- Comparison: if f(x) ≥ g(x) on the interval, then ∫ₐᵇ f dx ≥ ∫ₐᵇ g dx.
For example, since ∫₀² x² dx = 8/3, the constant multiple rule gives ∫₀² 3x² dx = 8, and reversing gives ∫₂⁰ x² dx = −8/3.
Section VAntiderivatives and the basic rules
An antiderivative of f is a function F with F′(x) = f(x). Since (x³)′ = 3x², x³ is an antiderivative of 3x². So is x³ + 5, and x³ − 2: adding a constant does not change a derivative. All the antiderivatives together are the indefinite integral:
∫ 3x² dx = x³ + CEach rule below is a derivative rule read backwards. You can always check an answer by differentiating it.
| Rule | Integral | Antiderivative | Note |
|---|---|---|---|
| Constant | ∫ k dx | kx + C | k is any constant. |
| Power | ∫ xⁿ dx | xⁿ⁺¹ over n + 1 + C | Any n ≠ −1: raise the power, divide by the new power. |
| Reciprocal | ∫ 1 over x dx | ln |x| + C | The n = −1 case the power rule misses. |
| Constant multiple | ∫ k f(x) dx | k ∫ f(x) dx | Constants slide outside. |
| Sum and difference | ∫ (f ± g) dx | ∫ f dx ± ∫ g dx | Integrate term by term. |
| Sine | ∫ sin x dx | −cos x + C | Check: (−cos x)′ = sin x. |
| Cosine | ∫ cos x dx | sin x + C | Check: (sin x)′ = cos x. |
| Exponential | ∫ eˣ dx | eˣ + C | eˣ is its own derivative. |
Section VIu-substitution: the chain rule in reverse
The chain rule says (F(g(x)))′ = F′(g(x)) g′(x). So when an integrand looks like something of an inside function, times the inside function’s derivative, rename the inside u = g(x). Then du = g′(x) dx and the integral becomes simpler.
For ∫ 2x(x² + 1)³ dx: let u = x² + 1, so du = 2x dx. The integral becomes ∫ u³ du = u⁴/4 + C, which is (x² + 1)⁴/4 + C.
The steps: choose u (usually the inside of a power, root or trig function), find du, rewrite everything in u, integrate, and put x back.
Section VIIA first look at the Fundamental Theorem
Riemann sums define the integral, but adding hundreds of rectangles by hand is slow. The surprise that links the two halves of calculus is this: if F is any antiderivative of a continuous function f, then
∫ₐᵇ f(x) dx = F(b) − F(a)Try it on the hero graph: F(x) = x³/3 gives F(2) − F(0) = 8/3, exactly the number the Riemann sums were closing in on. For sin x from 0 to π, F(x) = −cos x gives 1 − (−1) = 2. Why this works, and what it says about area as a function, is the next unit: the Fundamental Theorem of Calculus.
Section VIIIWorked examples
Example 1: four Riemann sums
Estimate ∫₀² x² dx with n = 4 using left, right, midpoint and trapezoid sums.
✓ Numerical check: the exact value is 8/3 ≈ 2.666667. The midpoint sum is off by 0.041667 and the trapezoid sum by 0.083333; set the explorer to n = 4 to see them.
Example 2: signed area
Find ∫₋₁² (x³ − 3x) dx and explain the sign.
✓ The answer is negative because more of the region from 0 to 2 lies below the axis than lies above it from −1 to 0. The explorer’s midpoint sum with n = 50 is −0.75135.
Example 3: an indefinite integral
Find ∫ (6x² − 4x + 5) dx.
✓ Check: (2x³ − 2x² + 5x)′ = 6x² − 4x + 5.
Example 4: u-substitution
Find ∫ 2x(x² + 1)³ dx.
✓ Check: ((x² + 1)⁴/4)′ = (x² + 1)³ · 2x. Numerically, from 0 to 1 the answer gives (16 − 1)/4 = 3.75, and a midpoint sum with n = 1000 gives 3.75000.
Example 5: using properties
Given ∫₀³ f(x) dx = 5 and ∫₀³ g(x) dx = −2, find ∫₀³ (2f(x) − 3g(x)) dx and ∫₃⁰ f(x) dx.
✓ Check with f(x) = 10x/9 and g(x) = −4x/9, which have those integrals: 2f − 3g = 32x/9, and its integral from 0 to 3 is 32 · 9/18 = 16.
Section IXCommon mistakes
- Forgetting + C. An indefinite integral is a whole family of functions.
- Using the power rule on 1/x. It would divide by zero. ∫ (1/x) dx = ln |x| + C.
- Treating signed area as total area. Parts below the axis subtract. For total area, integrate each part separately and add the sizes.
- Leaving x’s behind in u-substitution. Every x and the dx must be rewritten in u before you integrate.
- Thinking a left sum is always low. That is true only when the function is increasing.
Section XPractice problems
Try each one on paper first, then open the answer to check your steps. Every answer ends with a check you can repeat.
Estimate ∫₀⁴ x dx with a left-endpoint sum and n = 4. Left sum
Show answer and steps
Δx = 1; left endpoints 0, 1, 2, 3Four pieces of width 1L₄ = 1 · (0 + 1 + 2 + 3) = 6Answer✓ The exact area is a triangle, ½ · 4 · 4 = 8. The right sum is 10, and their average, 8, is the trapezoid sum.
Estimate ∫₀² x³ dx with a midpoint sum and n = 2. Midpoint sum
Show answer and steps
Δx = 1; midpoints 0.5 and 1.5Two pieces of width 1M₂ = 1 · (0.125 + 3.375) = 3.5Answer✓ The exact value is 2⁴/4 = 4, so the midpoint sum is 0.5 too small.
Estimate ∫₁² 1 over x dx with the trapezoid rule and n = 2. Trapezoid sum
Show answer and steps
Δx = 0.5; f(1) = 1, f(1.5) = ⅔, f(2) = 0.5Heights at the three pointsT₂ = 0.5 over 2(1 + 2 · ⅔ + 0.5) ≈ 0.708333Answer✓ The exact value is ln 2 ≈ 0.693147. The curve bends upward, so the trapezoids sit above it and overestimate.
Use geometry to find ∫₀³ (x − 2) dx. Signed area
Show answer and steps
Below the axis on 0 ≤ x ≤ 2: a triangle of area ½ · 2 · 2 = 2Counts as −2Above the axis on 2 ≤ x ≤ 3: a triangle of area ½ · 1 · 1 = ½Counts as +½∫₀³ (x − 2) dx = −2 + ½ = −1.5Answer✓ Check with an antiderivative: [x²/2 − 2x] from 0 to 3 = (4.5 − 6) − 0 = −1.5.
Find ∫ (3x² − 4x + 7) dx. Power rule
Show answer and steps
∫ 3x² dx = x³, ∫ 4x dx = 2x², ∫ 7 dx = 7xTerm by termx³ − 2x² + 7x + CAnswer✓ Differentiate to check: (x³ − 2x² + 7x)′ = 3x² − 4x + 7.
Find ∫ (√x + 1 over x) dx for x > 0. Mixed rules
Show answer and steps
√x = x1/2, so ∫ x1/2 dx = ⅔ x3/2Power rule with n = ½∫ 1 over x dx = ln xReciprocal rule⅔ x3/2 + ln x + CAnswer✓ Check: (⅔ x3/2)′ = x1/2 and (ln x)′ = 1/x.
Find ∫ x cos(x²) dx. u-substitution
Show answer and steps
u = x², du = 2x dx, so x dx = ½ duThe inside function and its derivative∫ ½ cos u du = ½ sin u + CIntegrate in u½ sin(x²) + CPut x back✓ Check: (½ sin(x²))′ = ½ cos(x²) · 2x = x cos(x²).
Find ∫ (2x + 3)⁵ dx. u-substitution
Show answer and steps
u = 2x + 3, du = 2 dx, so dx = ½ duLinear inside∫ ½ u⁵ du = u⁶ over 12 + CPower rule in u(2x + 3)⁶ over 12 + CAnswer✓ Check: the derivative is 6(2x + 3)⁵ · 2 / 12 = (2x + 3)⁵.
Use an antiderivative to find ∫₁³ (2x + 1) dx. FTC preview
Show answer and steps
F(x) = x² + xAn antiderivative of 2x + 1F(3) − F(1) = 12 − 2 = 10Answer✓ Check with geometry: the region is a trapezoid with parallel sides 3 and 7 and width 2, area (3 + 7)/2 · 2 = 10.
Section XISelf-guided quiz: eight quick questions
Choose one answer for each question, then press “Check my answers.” Your best score is saved only in this browser.
Section XIIFrom exhaustion to Riemann
Adding up thin pieces is one of the oldest ideas in mathematics. The Greek mathematician Eudoxus developed the method of exhaustion: fill a curved region with simpler shapes and squeeze the gap until it is smaller than any amount you name. Archimedes, around 225 BC, used it to show that a segment of a parabola has exactly 4/3 the area of the triangle with the same base and vertex, by fitting an endless sequence of ever-smaller triangles inside it.
Gottfried Wilhelm Leibniz gave integration its symbol. On 21 November 1675 he wrote a manuscript using the notation ∫ f(x) dx for the first time, the long S standing for a sum. He published his results on integration in 1684 and 1686 under the name calculus summatorius; the name “integral calculus” was suggested by Jacob Bernoulli in 1690.
Almost two centuries later, Bernhard Riemann made the idea precise. In his Habilitation dissertation at Göttingen, on representing functions by trigonometric series, he gave the conditions for a function to have an integral. That definition, built from the sums on this page, is still called the Riemann integral.
NotesSources and credits
- J. J. O’Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews: A history of the calculus, on Archimedes and the parabolic segment (around 225 BC), Leibniz’s ∫ notation, calculus summatorius and Jacob Bernoulli’s name “integral calculus” (1690).
- MacTutor: Eudoxus of Cnidus (biography), on the method of exhaustion.
- MacTutor: Archimedes (biography), on the method of exhaustion and Quadrature of the Parabola.
- MacTutor: Gottfried Wilhelm Leibniz (biography), on the manuscript of 21 November 1675.
- MacTutor: Bernhard Riemann (biography), on his Habilitation dissertation and the Riemann integral.
Credits: every graph and diagram on this page is original SVG, computed from the actual functions, made for Jugami. The interactive graph is drawn live in your browser with plain JavaScript, no outside libraries.