Question

Find the area between the graph of the given function and the x-axis over the given interval, if possible.

f(x)=1/(x-1)^2, for [-[infinity], 0]

Answer 1

`\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx = 1`

General Formulas and Concepts:

Calculus

Limits

• Limit Rule [Variable Direct Substitution]:
  `\displaystyle \lim_(x \to c) x = c`

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals
• Improper Integrals

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

Integration Rule [Fundamental Theorem of Calculus 1]:
`\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)`

U-Substitution

Area of a Region Formula:
`\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx`

Explanation:

Step 1: Define

`\displaystyle f(x) = (1)/((x - 1)^2) \\\left[ -\infty ,\ 0 \right]`

Step 2: Integrate Pt. 1

1. Substitute in variables [Area of a Region Formula]:
   `\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx`
2. [Integral] Rewrite [Improper Integral]:
   `\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx = \lim_(a \to -\infty) \int\limits^(0)_(a) {(1)/((x - 1)^2)} \, dx`

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

1. Set u:
   `\dipslaystyle u = x - 1`
2. [u] Basic Power Rule [Derivative Property - Addition/Subtraction]:
   `\dipslaystyle du = dx`
3. [Limits] Switch:
   `\displaystyle \left \{ {{x = 0 ,\ u = 0 - 1 = -1} \atop {x = a ,\ u = a - 1}} \right.`

Step 4: Integrate Pt. 3

1. [Integral] U-Substitution:
   `\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx = \lim_(a \to -\infty) \int\limits^(-1)_(a - 1) {(1)/(u^2)} \, du`
2. [Integral] Integration Rule [Reverse Power Rule]:
   `\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx = \lim_(a \to -\infty) (-1)/(x) \bigg| \limits^(-1)_(a - 1)`
3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx = \lim_(a \to -\infty) \bigg( (1)/(a - 1) + 1 \bigg)`
4. Evaluate limit [Limit Rule - Variable Direct Substitution]:
   `\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx = 0 + 1`
5. Simplify:
   `\displaystyle \int\limits^(0)_(- \infty) {(1)/((x - 1)^2)} \, dx = 1`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Advanced Integration Techniques