School curriculum· 40 min· 2 questions
Learn first: Derivatives and their applications
Integration is the inverse of differentiation: recovering a function from its known derivative. The definite integral lets us compute areas of curved figures and the distance travelled from a known velocity. Exams often ask for antiderivatives, the Newton–Leibniz formula and areas.
If F′(x) = f(x), then F is an antiderivative of f. For example, x³ is an antiderivative of 3x². But x³ + 5 and x³ − 2 are antiderivatives too, because the derivative of a constant is 0. So we write the whole family:
∫f(x)dx = F(x) + C
Properties:
∫(f ± g)dx = ∫f dx ± ∫g dx, ∫k·f dx = k∫f dx
Any answer can be checked by differentiating: the derivative of the result must equal the integrand.
Find the antiderivative of f(x) = 3x² − 2 whose graph passes through M(1; 3).
∫xⁿdx = xⁿ⁺¹/(n + 1) + C (n ≠ −1), ∫dx/x = ln|x| + C
∫sin x dx = −cos x + C, ∫cos x dx = sin x + C
∫dx/cos²x = tg x + C, ∫dx/sin²x = −ctg x + C
∫eˣdx = eˣ + C, ∫aˣdx = aˣ/ln a + C
Linear-argument rule (inner function kx + b):
∫f(kx + b)dx = F(kx + b)/k + C
Why divide by k? Differentiating F(kx + b) by the chain rule produces a factor k, which must be cancelled.
Samples:
Find ∫dx/(1 − cos x).
∫ₐᵇf(x)dx = F(b) − F(a)
A definite integral is a number (the constant C cancels). Properties:
∫ₐᵃf dx = 0; swapping the limits changes the sign of the integral; ∫ₐᵇf dx = ∫ₐᶜf dx + ∫ᶜᵇf dx
The integral of an odd function over a symmetric interval is 0: ∫₋₂² x³dx = 0.
Compute ∫₀²(3x² − 2x)dx.
More: ∫₀^(π/2) cos x dx = sin(π/2) − sin 0 = 1; ∫₁ᵉ dx/x = ln e − ln 1 = 1.
If f(x) ≥ 0, the area under the curve: S = ∫ₐᵇf(x)dx
Between two curves (f ≥ g): S = ∫ₐᵇ(f(x) − g(x))dx
Algorithm:
If the graph lies below Ox, the integral is negative — for the area take its absolute value. For example, the area under y = sin x on [0; π]: −cos π + cos 0 = 2.
Find the area of the region bounded by y = x² and y = x + 2.
If the velocity v(t) is known, the distance travelled during [t₁; t₂] is:
s = ∫ v(t)dt (from t₁ to t₂)
For example, with v(t) = 3t² + 2t the distance in the first 2 seconds is t³ + t² |₀² = 8 + 4 = 12 m.
A definite integral does not always have to be computed head-on: its properties shorten the work and prevent many mistakes. They follow directly from the Newton–Leibniz formula.
∫ₐᵇ(f ± g)dx = ∫ₐᵇf dx ± ∫ₐᵇg dx; ∫ₐᵇk·f dx = k·∫ₐᵇf dx
∫ₐᵃf dx = 0; swapping the limits changes the sign of the integral
∫ₐᵇf dx = ∫ₐᵐf dx + ∫ₘᵇf dx — splitting the interval into pieces
Note: even though the integral of an odd function over a symmetric interval is zero, the area between the graph and Ox is not. For area the rule from §4 still applies — the part below Ox is taken with a modulus.
Evaluate ∫₋₂²(x³ − 3x + 4)dx and ∫₀³|x − 1|dx.