Comprehensive Guide to Integrals

Definition: An integral represents the area under a curve, the accumulated value, or the antiderivative of a function. The definite integral of a function f(x) from a to b is denoted by ∫_a^b f(x) dx and represents the net area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.

1. Basic Integral Concepts

1.1 Indefinite Integrals

An indefinite integral is a family of functions whose derivative is the integrand. It is written as ∫ f(x) dx = F(x) + C, where F(x) is an antiderivative of f(x) and C is the constant of integration.

Example 1: Find ∫ x² dx

∫ x² dx = \frac{x^3}{3} + C

To verify, we can differentiate the result:

\frac{d}{dx}(\frac{x^3}{3} + C) = x^2, which matches our original function.

Example 2: Find ∫ (3x^2 + 2x - 4) dx

We can integrate each term separately:

\begin{align} ∫ (3x^2 + 2x - 4) dx &= 3∫ x^2 dx + 2∫ x dx - 4∫ dx \\ &= 3 \cdot \frac{x^3}{3} + 2 \cdot \frac{x^2}{2} - 4x + C \\ &= x^3 + x^2 - 4x + C \end{align}

1.2 Definite Integrals

A definite integral gives the net area between the function and the x-axis over a specific interval. It is written as ∫_a^b f(x) dx.

The fundamental theorem of calculus states that if F(x) is an antiderivative of f(x), then:

∫_a^b f(x) dx = F(b) - F(a)

Example 3: Evaluate ∫₀³ x^2 dx

\begin{align} ∫₀³ x^2 dx &= \left[ \frac{x^3}{3} \right]_0^3 \\ &= \frac{3^3}{3} - \frac{0^3}{3} \\ &= 9 - 0 \\ &= 9 \end{align}

Therefore, the definite integral equals 9 square units.

1.3 Basic Properties of Integrals

Here are some fundamental properties of integrals:

∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
∫ c \cdot f(x) dx = c \cdot ∫ f(x) dx, where c is a constant
∫_a^b f(x) dx = -∫_b^a f(x) dx
∫_a^b f(x) dx + ∫_b^c f(x) dx = ∫_a^c f(x) dx

1.4 Common Integration Formulas

Function	Integral
xⁿ (n ≠ -1)	\frac{x^{n+1}}{n+1} + C
\frac{1}{x} or x^-1	\ln\|x\| + C
e^x	e^x + C
\sin(x)	-\cos(x) + C
\cos(x)	\sin(x) + C
\tan(x)	-\ln\|\cos(x)\| + C or \ln\|\sec(x)\| + C
\sec^2(x)	\tan(x) + C
\frac{1}{1+x^2}	\arctan(x) + C
\frac{1}{\sqrt{1-x^2}}	\arcsin(x) + C

2. Integration Techniques

2.1 Substitution Method (u-substitution)

The substitution method (also known as u-substitution) is used when the integrand can be rewritten in terms of a new variable that simplifies the integration.

Steps:

Choose a substitution u = g(x)
Find du = g'(x) dx
Rewrite the integral in terms of u
Integrate with respect to u
Substitute back to express the answer in terms of x

Example 4: Evaluate ∫ x \cdot \cos(x^2) dx

Let u = x², then du = 2x dx

Rearranging: x dx = du/2

\begin{align} ∫ x \cdot \cos(x^2) dx &= ∫ \cos(u) \cdot \frac{du}{2} \\ &= \frac{1}{2} ∫ \cos(u) du \\ &= \frac{1}{2} \sin(u) + C \\ &= \frac{1}{2} \sin(x^2) + C \end{align}

2.2 Integration by Parts

Integration by parts is based on the product rule of differentiation and is useful for integrals of products of functions. The formula is:

∫ u(x) \cdot v'(x) dx = u(x) \cdot v(x) - ∫ u'(x) \cdot v(x) dx

Often written as: ∫ u \, dv = u \cdot v - ∫ v \, du

LIATE Rule for Choosing u: When deciding which function to choose as u, follow the LIATE order of preference:

L - Logarithmic functions (ln x, log x)
I - Inverse trigonometric functions (arcsin x, arctan x)
A - Algebraic functions (polynomials, roots)
T - Trigonometric functions (sin x, cos x)
E - Exponential functions (e^x, a^x)

Choose u as the function that comes first in this order.

Example 5: Find ∫ x \cdot e^x dx

Let u = x and dv = e^x dx
Then du = dx and v = e^x

\begin{align} ∫ x \cdot e^x dx &= x \cdot e^x - ∫ e^x dx \\ &= x \cdot e^x - e^x + C \\ &= e^x(x - 1) + C \end{align}

Example 6: Calculate ∫ \ln(x) dx

Let u = ln(x) and dv = dx
Then du = 1/x dx and v = x

\begin{align} ∫ \ln(x) dx &= x \cdot \ln(x) - ∫ x \cdot \frac{1}{x} dx \\ &= x \cdot \ln(x) - ∫ dx \\ &= x \cdot \ln(x) - x + C \\ &= x(\ln(x) - 1) + C \end{align}

2.3 Partial Fractions

Partial fraction decomposition is used to integrate rational functions (fractions with polynomials). The procedure involves breaking down a complex fraction into simpler fractions that are easier to integrate.

Example 7: Find ∫ \frac{5x+3}{x^2-1} dx

First, we factorize the denominator: x² - 1 = (x+1)(x-1)

Then, we decompose the fraction: \frac{5x+3}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1}

Multiply both sides by (x+1)(x-1): 5x+3 = A(x-1) + B(x+1)

Let's find A and B:

When x = 1: 5(1) + 3 = B(1+1) → 8 = 2B → B = 4

When x = -1: 5(-1) + 3 = A((-1)-1) → -2 = -2A → A = 1

So, \frac{5x+3}{x^2-1} = \frac{1}{x+1} + \frac{4}{x-1}

Now, we can integrate: \begin{align} ∫ \frac{5x+3}{x^2-1} dx &= ∫ \left(\frac{1}{x+1} + \frac{4}{x-1}\right) dx \\ &= ∫ \frac{1}{x+1} dx + 4∫ \frac{1}{x-1} dx \\ &= \ln|x+1| + 4\ln|x-1| + C \\ &= \ln|x+1| + \ln|x-1|^4 + C \\ &= \ln|x+1| \cdot |x-1|^4 + C \end{align}

2.4 Trigonometric Substitution

Trigonometric substitution is useful for integrals involving expressions like √(a²-x²), √(a²+x²), or √(x²-a²).

Trigonometric Substitution Guidelines:

For √(a²-x²), use x = a·sin(θ)
For √(a²+x²), use x = a·tan(θ)
For √(x²-a²), use x = a·sec(θ)

Example 8: Evaluate ∫ \frac{1}{\sqrt{4-x^2}} dx

This involves √(4-x²), so we use x = 2·sin(θ).

Then dx = 2·cos(θ) dθ and √(4-x²) = √(4-4sin²(θ)) = 2·cos(θ)

\begin{align} ∫ \frac{1}{\sqrt{4-x^2}} dx &= ∫ \frac{2\cos(\theta)}{2\cos(\theta)} d\theta \\ &= ∫ d\theta \\ &= \theta + C \\ &= \arcsin\left(\frac{x}{2}\right) + C \end{align}

We substitute back using θ = arcsin(x/2).

2.5 Integration Using Trigonometric Identities

For integrals involving products of sines and cosines, tangents and secants, etc., trigonometric identities can simplify the integration.

Example 9: Find ∫ \sin^2(x) dx

We use the identity sin²(x) = (1 - cos(2x))/2:

\begin{align} ∫ \sin^2(x) dx &= ∫ \frac{1 - \cos(2x)}{2} dx \\ &= \frac{1}{2}∫ dx - \frac{1}{2}∫ \cos(2x) dx \\ &= \frac{x}{2} - \frac{1}{2} \cdot \frac{\sin(2x)}{2} + C \\ &= \frac{x}{2} - \frac{\sin(2x)}{4} + C \\ &= \frac{x - \frac{\sin(2x)}{2}}{2} + C \end{align}

3. Applications of Integration

3.1 Area Between Curves

The area between two curves f(x) and g(x) from x = a to x = b, where f(x) ≥ g(x), is given by:

Area = ∫_a^b [f(x) - g(x)] dx

Example 10: Find the area between y = x² and y = x for 0 ≤ x ≤ 1.

For 0 ≤ x ≤ 1, we have x ≥ x². So, the area is:

\begin{align} Area &= ∫₀¹ [x - x^2] dx \\ &= ∫₀¹ (x - x^2) dx \\ &= \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 \\ &= \left( \frac{1}{2} - \frac{1}{3} \right) - \left( 0 - 0 \right) \\ &= \frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6} \end{align}

Therefore, the area between the curves is 1/6 square units.

3.2 Volume of Revolution

The volume of the solid obtained by rotating the region bounded by y = f(x), y = 0, x = a, and x = b around the x-axis is given by:

V = π ∫_a^b [f(x)]^2 dx

Example 11: Find the volume of the solid formed by rotating the region bounded by y = √x, y = 0, and x = 4 about the x-axis.

Using the formula for the volume of revolution about the x-axis:

\begin{align} V &= π ∫₀⁴ (\sqrt{x})^2 dx \\ &= π ∫₀⁴ x \, dx \\ &= π \left[ \frac{x^2}{2} \right]_0^4 \\ &= π \left( \frac{16}{2} - \frac{0}{2} \right) \\ &= 8π \end{align}

Therefore, the volume of the solid is 8π cubic units.

3.3 Arc Length

The length of a curve y = f(x) from x = a to x = b is given by:

Length = ∫_a^b \sqrt{1 + [f'(x)]^2} dx

Example 12: Find the arc length of y = x^3/2 from x = 0 to x = 1.

First, we find f'(x) = (3/2)x^1/2

\begin{align} Length &= ∫₀¹ \sqrt{1 + \left(\frac{3}{2}x^{1/2}\right)^2} dx \\ &= ∫₀¹ \sqrt{1 + \frac{9}{4}x} dx \end{align}

This is a complex integral. Let's use the substitution u = 1 + (9/4)x, which gives dx = (4/9) du, and when x = 0, u = 1, and when x = 1, u = 1 + 9/4 = 13/4.

\begin{align} Length &= ∫₁^13/4 \sqrt{u} \cdot \frac{4}{9} du \\ &= \frac{4}{9} ∫₁^13/4 u^{1/2} du \\ &= \frac{4}{9} \left[ \frac{2u^{3/2}}{3} \right]_1^{13/4} \\ &= \frac{4}{9} \cdot \frac{2}{3} \left[ u^{3/2} \right]_1^{13/4} \\ &= \frac{8}{27} \left[ \left(\frac{13}{4}\right)^{3/2} - 1^{3/2} \right] \\ &= \frac{8}{27} \left[ \frac{13\sqrt{13}}{8} - 1 \right] \\ &= \frac{8}{27} \left[ \frac{13\sqrt{13} - 8}{8} \right] \\ &= \frac{13\sqrt{13} - 8}{27} \end{align}

3.4 Average Value

The average value of a function f(x) over the interval [a, b] is given by:

f_avg = \frac{1}{b-a} ∫_a^b f(x) dx

Example 13: Find the average value of f(x) = sin(x) over the interval [0, π].

\begin{align} f_{avg} &= \frac{1}{π-0} ∫₀^π \sin(x) dx \\ &= \frac{1}{π} \left[ -\cos(x) \right]_0^π \\ &= \frac{1}{π} [(-\cos(π)) - (-\cos(0))] \\ &= \frac{1}{π} [(-(-1)) - (-1)] \\ &= \frac{1}{π} [1 + 1] \\ &= \frac{2}{π} \\ &≈ 0.637 \end{align}

3.5 Work Done by a Variable Force

The work done by a variable force F(x) in moving an object from x = a to x = b is given by:

Work = ∫_a^b F(x) dx

Example 14: A spring has a natural length of 10 cm. If it takes 8 N of force to stretch the spring to 14 cm, find the work done in stretching the spring from 12 cm to 16 cm.

Hooke's Law states that F = kx, where x is the displacement from the natural length.

First, we find the spring constant k: 8 = k(14 - 10) \\ 8 = 4k \\ k = 2 \text{ N/cm}

The force at displacement x from the natural length is F(x) = 2x, where x is in cm.

Now, we calculate the work done: \begin{align} Work &= ∫_12-10^16-10 2x \, dx \\ &= ∫₂⁶ 2x \, dx \\ &= 2 \left[ \frac{x^2}{2} \right]_2^6 \\ &= 2 \left[ \frac{36}{2} - \frac{4}{2} \right] \\ &= 2 [18 - 2] \\ &= 2 \cdot 16 \\ &= 32 \text{ N⋅cm} = 0.32 \text{ J} \end{align}