Answer:
[tex]\displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty)[/tex]
General Formulas and Concepts:
Pre-Calculus
Calculus
- Limits
- Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
- Integrals
- Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
- Trig Integration
- Improper Integrals
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^0_\infty {cos(x)} \, dx[/tex]
Step 2: Integrate
- [Improper Integral] Rewrite: [tex]\displaystyle \lim_{a \to \infty} \int\limits^0_a {cos(x)} \, dx[/tex]
- [Integral] Trig Integration: [tex]\displaystyle \lim_{a \to \infty} sin(x) \bigg| \limits^0_a[/tex]
- [Integral] Evaluate [Integration Rule - FTC 1]: [tex]\displaystyle \lim_{a \to \infty} sin(0) - sin(a)[/tex]
- Evaluate trig: [tex]\displaystyle \lim_{a \to \infty} -sin(a)[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle -sin(\infty)[/tex]
Since we are dealing with infinity of functions, we can do a numerous amount of things:
- Since -sin(x) is a shift from the parent graph sin(x), we can say that -sin(∞) = sin(∞) since sin(x) is an oscillating graph. The values of -sin(x) already have values in sin(x).
- Since sin(x) is an oscillating graph, we can also say that the integral actually equates to undefined, since it will never reach 1 certain value.
∴ [tex]\displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty) \ or \ \text{unde}\text{fined}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Improper Integrals
Book: College Calculus 10e
