Mathematics, 15.01.2020 05:31 mateoperez496

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]

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Answer from: faithbar0911p5p68a

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

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Differentiation

DerivativesDerivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

IntegralsImproper Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

$\displaystyle f(x) = \frac{1}{(x - 1)^2} \\\left[ -\infty ,\ 0 \right]$

Step 2: Integrate Pt. 1

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

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

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

Step 4: Integrate Pt. 3

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

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

Unit: Advanced Integration Techniques

Answer from: Quest

step-by-step explanation:

Answer from: Quest

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