- Find antiderivatives with the table and linearity
- Explain the definite integral as a limit of Riemann sums
- Evaluate definite integrals with the Newton–Leibniz formula
- Find the area between curves and the average value of a function
A car's trip computer records the speed every second. How far has the car gone? If the speed were constant, distance = speed × time, but the speed keeps changing. Split the trip into tiny time intervals, multiply speed by time in each, add up the products and let the intervals shrink: the result is an integral. Surprisingly, it is computed by running differentiation backwards.
Antiderivatives and the indefinite integral
F is an antiderivative of f on an interval if F′(x) = f(x) at every point of it. For example, x³/3 is an antiderivative of x², and so are x³/3 + 5 and x³/3 − π: any two antiderivatives differ by a constant.
- ∫the integral sign
- f(x)the integrand
- dxshows that x is the variable of integration
- Can arbitrary constant
The indefinite integral is the whole family of antiderivatives. Linearity: ∫ (a · f + b · g) dx = a ∫ f dx + b ∫ g dx.
| f(x) | ∫ f(x) dx | Note |
|---|---|---|
| xⁿ | xⁿ⁺¹ / (n + 1) + C | n ≠ −1; ∫ x³ dx = x⁴/4 + C |
| 1 / x | ln |x| + C | x ≠ 0; the case n = −1 |
| eˣ | eˣ + C | its own antiderivative |
| aˣ | aˣ / ln a + C | a > 0, a ≠ 1 |
| sin x | −cos x + C | mind the minus sign |
| cos x | sin x + C | x in radians |
| 1 / cos² x | tan x + C | cos x ≠ 0 |
| 1 / sin² x | −cot x + C | sin x ≠ 0 |
| 1 / (1 + x²) | arctan x + C | all x |
| 1 / √(1 − x²) | arcsin x + C | |x| < 1 |
| 1 / (x² + a²) | (1/a) arctan(x/a) + C | a ≠ 0 |
- Fan antiderivative of f
- k, bconstants, k ≠ 0
Linear-argument rule: ∫ cos 3x dx = (1/3) sin 3x + C, ∫ e²ˣ⁺¹ dx = (1/2) e²ˣ⁺¹ + C, ∫ dx/(2x − 5) = (1/2) ln|2x − 5| + C.
a) ∫ (3x² − 4/x + 2eˣ) dx
b) ∫ (√x + 1/x²) dx
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b) Write as powers: x^(1/2) + x⁻². ∫ x^(1/2) dx = x^(3/2)/(3/2) = (2/3)x^(3/2); ∫ x⁻² dx = x⁻¹/(−1) = −1/x.
Answer: (2/3)x√x − 1/x + C.
Check: ((2/3)x^(3/2))′ = x^(1/2) = √x and (−1/x)′ = 1/x². ✓
The definite integral: Riemann sums
Let f be continuous on [a, b]. Divide [a, b] into n equal parts of width Δx = (b − a)/n, pick a point xᵢ* in each part and form the Riemann sum ∑ f(xᵢ*) · Δx, the total area of n thin rectangles. As n → ∞ the sums approach a single number, the definite integral. For f ≥ 0 it is the area under the graph; where f < 0, the area counts with a minus sign (signed area).
- a, bthe lower and upper limits of integration
- xᵢ*a point chosen in the i-th subinterval
- Δxthe width of each subinterval
Definition of the definite integral (Riemann). The sign ∫ is a stretched S for “sum” (summa), and dx recalls Δx.
Estimate ∫₀¹ x² dx with n = 4 rectangles, using right and left endpoints.
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R₄ = (1/16 + 4/16 + 9/16 + 16/16) · 1/4 = 30/64 ≈ 0.469.
Left endpoints 0, 1/4, 1/2, 3/4: L₄ = (0 + 1/16 + 4/16 + 9/16) · 1/4 = 14/64 ≈ 0.219.
The exact area lies between them. With n rectangles Rₙ = (n + 1)(2n + 1)/(6n²) → 2/6 = 1/3 as n → ∞.
The Fundamental Theorem of Calculus
Adding up millions of rectangles is tedious. The great discovery of Newton and Leibniz links area to antiderivatives. Consider the area function A(x) = ∫ₐˣ f(t) dt. Increasing x by h adds a thin strip of width h and height about f(x): A(x + h) − A(x) ≈ f(x) · h. Divide by h and let h → 0: A′(x) = f(x). So the area function is an antiderivative of f!
- ∫ₐˣ f(t) dtthe area function A(x), an integral with a variable upper limit
- tthe integration variable (a different letter so it is not confused with x)
The Fundamental Theorem, part 1: differentiation undoes integration.
- Fany antiderivative of f
- F(x) |ₐᵇshort notation for F(b) − F(a)
The Newton–Leibniz formula (part 2 of the Fundamental Theorem). The constant C cancels in the difference, so any antiderivative will do.
a) ∫₀¹ x² dx b) ∫₀^π sin x dx c) ∫₁^e dx/x d) ∫₀² (3x² − 2x + 1) dx
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b) [−cos x]₀^π = −cos π + cos 0 = 1 + 1 = 2. One arch of the sine curve has area 2.
c) [ln x]₁^e = ln e − ln 1 = 1.
d) [x³ − x² + x]₀² = 8 − 4 + 2 = 6.
- Swapping the limits changes the sign; ∫ₐᵃ f dx = 0.
- Additivity: ∫ₐᵇ f dx = ∫ₐᵐ f dx + ∫ₘᵇ f dx for any m.
- Linearity: ∫ₐᵇ (αf + βg) dx = α ∫ₐᵇ f dx + β ∫ₐᵇ g dx.
- Comparison: if f(x) ≤ g(x) on [a, b], then ∫ₐᵇ f dx ≤ ∫ₐᵇ g dx.
- Symmetry: for odd f, ∫₋ₐᵃ f dx = 0; for even f, ∫₋ₐᵃ f dx = 2 ∫₀ᵃ f dx.
Area between curves and average value
- fthe upper curve: f(x) ≥ g(x) on [a, b]
- gthe lower curve
- a, busually the x-coordinates where the curves intersect
Area between two curves: “upper minus lower”, then integrate.
Find the area of the region enclosed by y = x and y = x².
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On [0, 1], x ≥ x² (e.g. at x = 1/2: 1/2 > 1/4), so the line is on top.
S = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
- f̄the average value of f on [a, b]
- b − athe length of the interval
Mean value theorem for integrals: a continuous f actually takes the value f̄ at some point c of the interval. The rectangle of height f̄ has the same area as the region under the graph.
a) Find the average value of sin x on [0, π].
b) A body moves with speed v(t) = 3t² m/s. Find the distance travelled for 0 ≤ t ≤ 2 s and the average speed.
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b) Distance: s = ∫₀² 3t² dt = [t³]₀² = 8 m.
Average speed: 8/2 = 4 m/s. Check: the speed grows from 0 to 12 m/s, and the average lies in between.
Key points
- If F′ = f, then F is an antiderivative of f, and ∫ f dx = F + C.
- ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1), ∫ dx/x = ln|x| + C, ∫ eˣ dx = eˣ + C.
- The definite integral is the limit of Riemann sums: the signed area under the graph.
- Newton–Leibniz: ∫ₐᵇ f dx = F(b) − F(a); also d/dx ∫ₐˣ f(t) dt = f(x).
- Area between curves: ∫ₐᵇ (upper − lower) dx; average value: (1/(b − a)) ∫ₐᵇ f dx.
Check yourself
10 questions. Every correct answer earns XP.