Respuesta :
Answer:
[tex] \int_{1}^{\infty} \frac{1}{x} dx = \infty -0 [/tex]
And with that we see that our integral diverges.
Step-by-step explanation:
For this case we assume the following integral:
[tex] \int_{1}^{\infty} \frac{1}{x} dx[/tex]
Since 1.000 = 1 are equivalent. We can solve the integral and since we know that [tex] \int \frac{1}{x} dx = ln|x| +c [/tex] we got this:
[tex] \int_{1}^{\infty} \frac{1}{x} dx = ln |x| \Big|_1^{\infty} [/tex]
And when we evaluate the limits using the fundamental theorem of calculus we got:
[tex] \int_{1}^{\infty} \frac{1}{x} dx = \lim_{x\to\infty} ln|x| - ln |1| [/tex]
By properties we know that [tex] ln |1|=0[/tex]. But the [tex]\lim_{x\to\infty} ln|x|[/tex] is not defined since the natural log is a continuous increasing function so then:
[tex] \lim_{x\to\infty} ln|x| = \infty[/tex]
And if we repplace this into our integral we got:
[tex] \int_{1}^{\infty} \frac{1}{x} dx = \infty -0 [/tex]
And with that we see that our integral diverges.
Answer:
The improper integral diverges.
[tex]\displaystyle \int\limits^{\infty}_1 {\frac{1}{x}} \, dx = \infty[/tex]
General Formulas and Concepts:
Calculus
Limit
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Integration
- Integrals
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Improper Integral: [tex]\displaystyle \int\limits^{\infty}_a {f(x)} \, dx = \lim_{b \to \infty} \int\limits^b_a {f(x)} \, dx[/tex]
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle \int\limits^{\infty}_1 {\frac{1}{x}} \, dx[/tex]
Step 2: Integrate
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{\infty}_1 {\frac{1}{x}} \, dx = \lim_{b \to \infty} \int\limits^{b}_1 {\frac{1}{x}} \, dx[/tex]
- [Integral] Apply Logarithmic Integration: [tex]\displaystyle \int\limits^{\infty}_1 {\frac{1}{x}} \, dx = \lim_{b \to \infty} \ln |x| \bigg| \limits^{b}_1[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\infty}_1 {\frac{1}{x}} \, dx = \lim_{b \to \infty} \ln |b|[/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{\infty}_1 {\frac{1}{x}} \, dx = \ln |\infty|[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_1 {\frac{1}{x}} \, dx = \infty[/tex]
∴ the improper integral tends to ∞ and is divergent.
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Learn more about improper integrals: https://brainly.com/question/14412498
Learn more about calculus: https://brainly.com/question/23558817
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
