Integral as (Signed) Area - If f f f is continuous and positive on [ a , b ] [a,b] [ a , b ] , ∫ a b f ( t ) d t \int_a^b f(t)\,dt ∫ a b f ( t ) d t is the area under the graph between x = a x=a x = a and x = b x=b x = b .
- If f f f changes sign, the integral is the algebraic sum: positive areas where f>0, negative where f<0.
Basic Properties - Zero interval: ∫ a a f ( t ) d t = 0. \int_a^a f(t)\,dt = 0. ∫ a a f ( t ) d t = 0.
- Orientation: ∫ a b f ( t ) d t = − ∫ b a f ( t ) d t . \int_a^b f(t)\,dt = -\int_b^a f(t)\,dt. ∫ a b f ( t ) d t = − ∫ b a f ( t ) d t .
- Chasles’ relation: for c ∈ [ a , b ] c\in[a,b] c ∈ [ a , b ] , ∫ a b f ( t ) d t = ∫ a c f ( t ) d t + ∫ c b f ( t ) d t . \int_a^b f(t)\,dt = \int_a^c f(t)\,dt + \int_c^b f(t)\,dt. ∫ a b f ( t ) d t = ∫ a c f ( t ) d t + ∫ c b f ( t ) d t .
- Linearity: for continuous f , g f,g f , g on [ a , b ] [a,b] [ a , b ] and λ ∈ R \lambda\in\mathbb{R} λ ∈ R ,
- ∫ a b ( f + g ) ( t ) d t = ∫ a b f ( t ) d t + ∫ a b g ( t ) d t , \int_a^b (f+g)(t)\,dt = \int_a^b f(t)\,dt + \int_a^b g(t)\,dt, ∫ a b ( f + g ) ( t ) d t = ∫ a b f ( t ) d t + ∫ a b g ( t ) d t ,
- ∫ a b λ f ( t ) d t = λ ∫ a b f ( t ) d t . \int_a^b \lambda f(t)\,dt = \lambda \int_a^b f(t)\,dt. ∫ a b λ f ( t ) d t = λ ∫ a b f ( t ) d t .
- Positivity / comparison:
- If f ≥ 0 f\ge 0 f ≥ 0 on [ a , b ] [a,b] [ a , b ] , then ∫ a b f ( t ) d t ≥ 0. \int_a^b f(t)\,dt \ge 0. ∫ a b f ( t ) d t ≥ 0.
- If f ≤ g f\le g f ≤ g on [ a , b ] [a,b] [ a , b ] , then ∫ a b f ( t ) d t ≤ ∫ a b g ( t ) d t . \int_a^b f(t)\,dt \le \int_a^b g(t)\,dt. ∫ a b f ( t ) d t ≤ ∫ a b g ( t ) d t .
#Primitive (Antiderivative) - A primitive of f f f on an interval I I I is a function F F F with F ′ ( x ) = f ( x ) F'(x)=f(x) F ′ ( x ) = f ( x ) for all x ∈ I x\in I x ∈ I .
- If F F F is one primitive, then all primitives of f f f are F + C F+C F + C with C ∈ R C\in\mathbb{R} C ∈ R .
- The function F ( x ) = ∫ a x f ( t ) d t F(x)=\int_a^x f(t)\,dt F ( x ) = ∫ a x f ( t ) d t is a primitive of f f f on [ a , b ] [a,b] [ a , b ] and satisfies F ( a ) = 0 F(a)=0 F ( a ) = 0 (distinguished primitive).
- Fundamental theorem of calculus: for continuous f f f on [ a , b ] [a,b] [ a , b ] and any primitive F F F , ∫ a b f ( t ) d t = F ( b ) − F ( a ) = [ F ( x ) ] a b . \int_a^b f(t)\,dt = F(b) - F(a) = [F(x)]_a^b. ∫ a b f ( t ) d t = F ( b ) − F ( a ) = [ F ( x ) ] a b .
Usual Primitives - Power functions (ε ≠ − 1 \varepsilon\neq -1 ε = − 1 ): ∫ x ε d x = x ε + 1 ε + 1 + C . \int x^\varepsilon\,dx = \frac{x^{\varepsilon+1}}{\varepsilon+1} + C. ∫ x ε d x = ε + 1 x ε + 1 + C .
- Logarithm case: ∫ 1 x d x = ln ∣ x ∣ + C . \int \frac{1}{x}\,dx = \ln|x| + C. ∫ x 1 d x = ln ∣ x ∣ + C .
- Exponential: ∫ e x d x = e x + C . \int e^x\,dx = e^x + C. ∫ e x d x = e x + C .
- Trigonometric: ∫ cos x d x = sin x + C , ∫ sin x d x = − cos x + C . \int \cos x\,dx = \sin x + C,\qquad \int \sin x\,dx = -\cos x + C. ∫ cos x d x = sin x + C , ∫ sin x d x = − cos x + C .
- Arctangent: ∫ 1 1 + x 2 d x = arctan x + C . \int \frac{1}{1+x^2}\,dx = \arctan x + C. ∫ 1 + x 2 1 d x = arctan x + C .
Integration by Parts - Based on ( f g ) ′ = f ′ g + f g ′ . (fg)' = f'g + fg'. ( f g ) ′ = f ′ g + f g ′ .
- Primitive form: ∫ f ′ ( x ) g ( x ) d x = f ( x ) g ( x ) − ∫ f ( x ) g ′ ( x ) d x . \int f'(x)g(x)\,dx = f(x)g(x) - \int f(x)g'(x)\,dx. ∫ f ′ ( x ) g ( x ) d x = f ( x ) g ( x ) − ∫ f ( x ) g ′ ( x ) d x .
- Definite integral form: ∫ a b f ′ ( x ) g ( x ) d x = [ f ( x ) g ( x ) ] a b − ∫ a b f ( x ) g ′ ( x ) d x . \int_a^b f'(x)g(x)\,dx = [f(x)g(x)]_a^b - \int_a^b f(x)g'(x)\,dx. ∫ a b f ′ ( x ) g ( x ) d x = [ f ( x ) g ( x ) ] a b − ∫ a b f ( x ) g ′ ( x ) d x .
- Typical strategy: choose g g g to simplify when differentiated and f ′ f' f ′ to be easily integrable.
# Example: Primitive of ln x \ln x ln x
- On ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ ) : ∫ ln x d x = x ln x − x + C , \int \ln x\,dx = x\ln x - x + C, ∫ ln x d x = x ln x − x + C , via integration by parts with f ′ ( x ) = 1 f'(x)=1 f ′ ( x ) = 1 , g ( x ) = ln x g(x)=\ln x g ( x ) = ln x .
Change of Variables (Substitution) - General theorem: if u : [ a , b ] → I u:[a,b]\to I u : [ a , b ] → I is C 1 \mathcal{C}^1 C 1 with continuous derivative and f f f is continuous on I I I , then ∫ a b f ( u ( t ) ) u ′ ( t ) d t = ∫ u ( a ) u ( b ) f ( x ) d x . \int_a^b f(u(t))u'(t)\,dt = \int_{u(a)}^{u(b)} f(x)\,dx. ∫ a b f ( u ( t )) u ′ ( t ) d t = ∫ u ( a ) u ( b ) f ( x ) d x .
- Careful: the bounds must also be transformed: a , b ↦ u ( a ) , u ( b ) a,b \mapsto u(a),u(b) a , b ↦ u ( a ) , u ( b ) .
- Primitive version: ∫ f ( u ( x ) ) u ′ ( x ) d x = F ( u ( x ) ) + C , \int f(u(x))u'(x)\,dx = F(u(x)) + C, ∫ f ( u ( x )) u ′ ( x ) d x = F ( u ( x )) + C , where F F F is a primitive of f f f .
Translation / Affine Change of Variable - Translation: ∫ a b f ( x + x 0 ) d x = ∫ a + x 0 b + x 0 f ( t ) d t . \int_a^b f(x+x_0)\,dx = \int_{a+x_0}^{b+x_0} f(t)\,dt. ∫ a b f ( x + x 0 ) d x = ∫ a + x 0 b + x 0 f ( t ) d t .
- General affine change u ( x ) = p x + x 0 u(x)=px+x_0 u ( x ) = p x + x 0 (p ≠ 0 p\neq0 p = 0 ):
- Primitive: ∫ f ( p x + x 0 ) d x = 1 p F ( p x + x 0 ) + C , \int f(px+x_0)\,dx = \frac{1}{p}F(px+x_0)+C, ∫ f ( p x + x 0 ) d x = p 1 F ( p x + x 0 ) + C , where F ′ = f F' = f F ′ = f .
- Definite integral: ∫ a b f ( p x + x 0 ) d x = 1 p ∫ p a + x 0 p b + x 0 f ( x ) d x . \int_a^b f(px+x_0)\,dx = \frac{1}{p}\int_{pa+x_0}^{pb+x_0} f(x)\,dx. ∫ a b f ( p x + x 0 ) d x = p 1 ∫ p a + x 0 p b + x 0 f ( x ) d x .
“Physicist’s Method” Notation - Write t = u ( x ) t = u(x) t = u ( x ) , compute d t d x \dfrac{dt}{dx} d x d t , rearrange symbolically as d x = ⋯ d t dx = \cdots\,dt d x = ⋯ d t to do substitution quickly.
- Same idea as rigorous change of variable, but with differential notation.
Example Types - Integral of t e − t t e^{-t} t e − t on [ 0 , T ] [0,T] [ 0 , T ] by parts; shows a limit as T → + ∞ T\to+\infty T → + ∞ equal to 1 1 1 (generalized integral): ∫ 0 T t e − t d t . \int_0^T t e^{-t}\,dt. ∫ 0 T t e − t d t .
- Substitution examples such as ∫ 1 9 1 1 + x d x , \int_1^9 \frac{1}{1+\sqrt{x}}\,dx, ∫ 1 9 1 + x 1 d x , with t = x t=\sqrt{x} t = x .
- Completing the square and substitution for rational functions like ∫ d x x 2 + 4 x + 8 . \int \frac{dx}{x^2+4x+8}. ∫ x 2 + 4 x + 8 d x .
Applied Example – River Discharge - A discharge function q ( t ) q(t) q ( t ) models water flow rate (e.g. in millions of m 3 \text{m}^3 m 3 per day).
- Total volume over [ 0 , T ] [0,T] [ 0 , T ] is v ( T ) = ∫ 0 T q ( t ) d t . v(T) = \int_0^T q(t)\,dt. v ( T ) = ∫ 0 T q ( t ) d t .
- In the example, q ( t ) q(t) q ( t ) is given in terms of a logistic-type expression, and recognizing it as q ( t ) = c − u ′ ( t ) u ( t ) 2 q(t) = c\,\frac{-u'(t)}{u(t)^2} q ( t ) = c u ( t ) 2 − u ′ ( t ) allows finding a primitive analytically.