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Higher Mathematics

Integral Calculus of One Variable

This chapter systematically studies indefinite and definite integrals of single-variable functions, including the basic concepts of integration, common integration methods, properties and applications of integrals, and their practical applications in geometry and physics.

Indefinite Integrals

Definition of Indefinite Integral

If F(x)F(x) is a derivative of f(x)f(x), then F(x)F(x) is called an antiderivative of f(x)f(x), and all antiderivatives of f(x)f(x) are denoted as

f(x)dx=F(x)+C\int f(x) dx = F(x) + C

where CC is a constant.

Basic Integration Formulas

  • xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C (n1n \neq -1)
  • 1xdx=lnx+C\int \frac{1}{x} dx = \ln|x| + C
  • exdx=ex+C\int e^{x} dx = e^{x} + C
  • sinxdx=cosx+C\int \sin x dx = -\cos x + C
  • cosxdx=sinx+C\int \cos x dx = \sin x + C

Properties of Indefinite Integrals

  • Linearity: [af(x)+bg(x)]dx=af(x)dx+bg(x)dx\int [af(x) + bg(x)] dx = a\int f(x) dx + b\int g(x) dx

Substitution Method

Let u=φ(x)u = \varphi(x), then

f(φ(x))φ(x)dx=f(u)du\int f(\varphi(x)) \varphi'(x) dx = \int f(u) du

Integration by Parts

udv=uvvdu\int u dv = uv - \int v du

Definite Integrals

Definition of Definite Integral

If f(x)f(x) is continuous on [a,b][a, b], then

abf(x)dx=limni=1nf(xi)Δx\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x

Properties of Definite Integrals

  • Linearity, additivity over intervals, sign-preserving property

Newton-Leibniz Formula

If F(x)F(x) is an antiderivative of f(x)f(x), then

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

Integral with Variable Upper Limit

F(x)=axf(t)dtF(x) = \int_a^x f(t) dt, F(x)=f(x)F'(x) = f(x)

Improper Integrals

  • If the interval of integration is unbounded or the integrand is unbounded in the interval, it is called an improper integral.

Integrals of Common Functions

  • Rational function integrals: partial fraction decomposition
  • Trigonometric function integrals: trigonometric identities, substitution
  • Irrational function integrals: substitution method

Applications of Definite Integrals

  • Area of a plane figure: S=abf(x)dxS = \int_a^b |f(x)| dx
  • Arc length of a plane curve: L=ab1+[f(x)]2dxL = \int_a^b \sqrt{1 + [f'(x)]^2} dx
  • Volume of a solid of revolution: V=πab[f(x)]2dxV = \pi \int_a^b [f(x)]^2 dx
  • Physical applications: work, center of mass, etc.

Exercises

  1. Compute the indefinite integral (2x33x2+1)dx\int (2x^3 - 3x^2 + 1) dx.
  2. Compute the definite integral 01xex2dx\int_0^1 x e^{x^2} dx.
  3. Use integration by parts to compute xcosxdx\int x \cos x dx.
  4. Compute the improper integral 11x2dx\int_1^{\infty} \frac{1}{x^2} dx.
  5. Find the area enclosed by the curve y=x2y = x^2 and the xx-axis on [0,1][0, 1].
  6. Compute the arc length of y=xy = \sqrt{x} on [0,4][0, 4].
Reference Answers

1. Compute the indefinite integral (2x33x2+1)dx\int (2x^3 - 3x^2 + 1) dx

Solution: Integrate term by term.

2x3dx=24x4=12x4\int 2x^3 dx = \frac{2}{4}x^4 = \frac{1}{2}x^4

3x2dx=x3\int -3x^2 dx = -x^3

1dx=x\int 1 dx = x

Answer: 12x4x3+x+C\frac{1}{2}x^4 - x^3 + x + C


2. Compute the definite integral 01xex2dx\int_0^1 x e^{x^2} dx

Solution: Substitute u=x2,du=2xdxu = x^2, du = 2x dx.

01xex2dx=1201eudu=12(e1e0)=12(e1)\int_0^1 x e^{x^2} dx = \frac{1}{2} \int_0^1 e^{u} du = \frac{1}{2}(e^1 - e^0) = \frac{1}{2}(e - 1)

Answer: 12(e1)\frac{1}{2}(e - 1)


3. Use integration by parts to compute xcosxdx\int x \cos x dx

Solution: Let u=x,dv=cosxdxu = x, dv = \cos x dx, du=dx,v=sinxdu = dx, v = \sin x

xcosxdx=xsinxsinxdx=xsinx+cosx+C\int x \cos x dx = x \sin x - \int \sin x dx = x \sin x + \cos x + C

Answer: xsinx+cosx+Cx \sin x + \cos x + C


4. Compute the improper integral 11x2dx\int_1^{\infty} \frac{1}{x^2} dx

Solution: 1Ax2dx=[x1]1A=(1A+1)\int_1^A x^{-2} dx = [-x^{-1}]_1^A = ( -\frac{1}{A} + 1 ), as AA \to \infty, get 11.

Answer: 11


5. Find the area enclosed by y=x2y = x^2 and the xx-axis on [0,1][0, 1]

Solution: S=01x2dx=13S = \int_0^1 x^2 dx = \frac{1}{3}

Answer: 13\frac{1}{3}


6. Compute the arc length of y=xy = \sqrt{x} on [0,4][0, 4]

Solution: L=041+[f(x)]2dxL = \int_0^4 \sqrt{1 + [f'(x)]^2} dx, f(x)=12x1/2f'(x) = \frac{1}{2}x^{-1/2}

1+[f(x)]2=1+14x1 + [f'(x)]^2 = 1 + \frac{1}{4x}

L=041+14xdxL = \int_0^4 \sqrt{1 + \frac{1}{4x}} dx

(You can leave it as is, or further simplify using substitution)

Answer: L=041+14xdxL = \int_0^4 \sqrt{1 + \frac{1}{4x}} dx (can be further simplified)