Integrals

Integrals, a fundamental concept in calculus, reverse differentiation to compute areas under curves or accumulated quantities like distance, volume, or work. Definite integrals calculate exact areas between points, while indefinite integrals yield antiderivatives. This MathMultiverse guide covers definitions, rules, step-by-step examples, and applications in physics, economics, and more, enhanced with interactive visualizations for intuitive learning.

Why are integrals vital? They solve real-world problems involving accumulation, from calculating total distance traveled to determining consumer surplus, making them essential for mathematics and applied sciences.

Definition

The definite integral \( \int_a^b f(x) \, dx \) computes the signed area under \( y = f(x) \) from \( x = a \) to \( x = b \), using the antiderivative \( F(x) \) where \( F'(x) = f(x) \):

\[ \int_a^b f(x) \, dx = F(b) - F(a) \]

The indefinite integral \( \int f(x) \, dx = F(x) + C \) represents the family of antiderivatives, with \( C \) as the constant of integration. Integrals model cumulative effects, such as total mass or energy, across various domains.

Examples

Let’s compute key integrals, both indefinite and definite, with interpretations.

1. \( \int x \, dx \)

Using the power rule (\( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), \( n \neq -1 \)):

\[ \int x \, dx = \frac{x^2}{2} + C \]

For \( \int_0^2 x \, dx \):

\[ \int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = 2 \]

This is the area of a triangle (base 2, height 2) under \( y = x \).

2. \( \int x^2 \, dx \)

Power rule:

\[ \int x^2 \, dx = \frac{x^3}{3} + C \]

For \( \int_1^3 x^2 \, dx \):

\[ \int_1^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \approx 8.67 \]

Area under the parabola \( y = x^2 \).

3. \( \int \frac{1}{x} \, dx \)

Special rule:

\[ \int \frac{1}{x} \, dx = \ln|x| + C \]

For \( \int_1^2 \frac{1}{x} \, dx \):

\[ \int_1^2 \frac{1}{x} \, dx = \left[ \ln x \right]_1^2 = \ln 2 - \ln 1 = \ln 2 \approx 0.693 \]

Area under the hyperbola \( y = \frac{1}{x} \).

4. \( \int \sqrt{x} \, dx \)

Rewrite \( \sqrt{x} = x^{1/2} \):

\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{\frac{3}{2}} + C \]

For \( \int_0^4 \sqrt{x} \, dx \):

\[ \int_0^4 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{\frac{3}{2}} \right]_0^4 = \frac{2}{3} (8) - 0 = \frac{16}{3} \approx 5.33 \]

Area under \( y = \sqrt{x} \).

5. \( \int x^3 \, dx \)

Power rule:

\[ \int x^3 \, dx = \frac{x^4}{4} + C \]

For \( \int_{-1}^1 x^3 \, dx \):

\[ \int_{-1}^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-1}^1 = \frac{1}{4} - \frac{1}{4} = 0 \]

Zero due to symmetry of \( y = x^3 \).

6. \( \int 5 \, dx \)

Constant rule:

\[ \int 5 \, dx = 5x + C \]

For \( \int_0^3 5 \, dx \):

\[ \int_0^3 5 \, dx = \left[ 5x \right]_0^3 = 15 - 0 = 15 \]

Area of a rectangle (height 5, width 3).

Graphical View

Visualizing integrals as areas under curves.

Area under \( y = x \) from 0 to 2

Triangle with area 2.

Area under \( y = x^2 \) from 1 to 3

Area \( \frac{26}{3} \).

Area under \( y = \sqrt{x} \) from 0 to 4

Area \( \frac{16}{3} \).

Rules

Key integration rules for finding antiderivatives.

Power Rule:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 \]

Examples: \( \int x^2 \, dx = \frac{x^3}{3} + C \), \( \int x^{-2} \, dx = -\frac{1}{x} + C \).
Constant Rule:
\[ \int k \, dx = kx + C \]

Example: \( \int 7 \, dx = 7x + C \).
Logarithmic Rule:
\[ \int \frac{1}{x} \, dx = \ln|x| + C \]

Example: \( \int_2^4 \frac{1}{x} \, dx = \ln 2 \).
Sum Rule:
\[ \int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx \]

Example: \( \int (x + x^2) \, dx = \frac{x^2}{2} + \frac{x^3}{3} + C \).
Constant Multiple Rule:
\[ \int k f(x) \, dx = k \int f(x) \, dx \]

Example: \( \int 3x^2 \, dx = x^3 + C \).

Applications

Integrals model accumulation in real-world scenarios.

Physics - Distance: Velocity \( v(t) = 3t^2 + 2 \), \( t = 0 \) to \( 2 \):
\[ s(t) = \int_0^2 (3t^2 + 2) \, dt = \left[ t^3 + 2t \right]_0^2 = 12 \]
Volume: Rotate \( y = x^2 \), \( x = 0 \) to \( 1 \), about x-axis:
\[ V = \pi \int_0^1 x^4 \, dx = \pi \left[ \frac{x^5}{5} \right]_0^1 = \frac{\pi}{5} \]
Work: Force \( F(x) = 2x \), \( x = 0 \) to \( 3 \):
\[ W = \int_0^3 2x \, dx = \left[ x^2 \right]_0^3 = 9 \]
Economics - Consumer Surplus: Demand \( p = 10 - x \), price 5:
\[ \text{CS} = \int_0^5 (5 - x) \, dx = \left[ 5x - \frac{x^2}{2} \right]_0^5 = 12.5 \]
Probability - Expected Value: Density \( f(x) = x \), \( x = 0 \) to \( 1 \):
\[ E(X) = \int_0^1 x^2 \, dx = \frac{1}{3} \]