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sweet [91]
2 years ago
6

\int\limits^0_∞ cos{x} \, dx

Mathematics
1 answer:
JulijaS [17]2 years ago
7 0

Answer:

\displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty)

General Formulas and Concepts:

<u>Pre-Calculus</u>

  • Unit Circle
  • Trig Graphs

<u>Calculus</u>

  • Limits
  • Limit Rule [Variable Direct Substitution]:                                                     \displaystyle \lim_{x \to c} x = c
  • Integrals
  • Integration Rule [Fundamental Theorem of Calculus 1]:                             \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)
  • Trig Integration
  • Improper Integrals

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^0_\infty {cos(x)} \, dx

<u>Step 2: Integrate</u>

  1. [Improper Integral] Rewrite:                                                                         \displaystyle  \lim_{a \to \infty} \int\limits^0_a {cos(x)} \, dx
  2. [Integral] Trig Integration:                                                                             \displaystyle  \lim_{a \to \infty} sin(x) \bigg| \limits^0_a
  3. [Integral] Evaluate [Integration Rule - FTC 1]:                                               \displaystyle  \lim_{a \to \infty} sin(0) - sin(a)
  4. Evaluate trig:                                                                                                 \displaystyle  \lim_{a \to \infty} -sin(a)
  5. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle  -sin(\infty)

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.

∴  \displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty) \ or \ \text{unde}\text{fined}

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

Unit: Improper Integrals

Book: College Calculus 10e

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3 years ago
3x + 1 = 5X-13 if anyone knows what to do can they please help ​
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Answer:

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Step-by-step explanation:

First, you look at the equation. Identify the locations of variable terms (both sides of the equal sign) and the constant terms (both sides of the equal sign).

If there are any parentheses, it is a good idea to use the distributive property to eliminate them. Here, there are none.

I like to start by subtracting <em>the variable term with the smallest coefficient</em>. Here, that is 3x, so we add -3x to both sides of the equation.

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Now, we have the only variable term on one side of the equal sign. We want it by itself, so we need to make the -13 go away. We do that by adding its opposite to both sides of the equation:

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__

It is generally a good idea to <em>check your work</em> by seeing if your solution value satisfies the equation:

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_____

<em>Additional comment</em>

By subtracting 3x from 5x, the result is 2x with a positive coefficient. We could solve the equation just as easily by subtracting 5x from 3x. That result would be ...

  -2x +1 = -13

Subtracting 1 would give

  -2x = -14

and you would multiply by -1/2 to get x=7. I personally like to avoid having this many minus signs show up in the problem. That is why I choose to subtract the x-term with the smallest coefficient.

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