Answer:
[tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \infty[/tex]
General Formulas and Concepts:
Algebra I
- Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Limit Property [Multiplied Constant]: [tex]\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)[/tex]
Integrals
- Definite Integrals
- Improper Integrals
Integration Constant C
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]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx[/tex]
Step 2: Integrate
- [Integrand] Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \int\limits^{\infty}_{\frac{\pi}{2}} {7x^{\frac{-1}{2}}} \, dx[/tex]
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = 7\int\limits^{\infty}_{\frac{\pi}{2}} {x^{\frac{-1}{2}}} \, dx[/tex]
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \lim_{b \to \infty} 7\int\limits^b_{\frac{\pi}{2}} {x^{\frac{-1}{2}}} \, dx[/tex]
- [Integral] Reverse Power Rule: [tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \lim_{b \to \infty} 7(2x^{\frac{1}{2}}) \bigg| \limits^b_{\frac{\pi}{2}}[/tex]
- [Limit] Rewrite [Limit Property - Multiplied Constant]: [tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = 7\lim_{b \to \infty} (2x^{\frac{1}{2}}) \bigg| \limits^b_{\frac{\pi}{2}}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = 7\lim_{b \to \infty} (2b^{\frac{1}{2}} - \sqrt{2\pi})[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \infty[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
