Mathematics, 23.01.2020 08:31 Casmereee

Calculus question, just the first line. other stuff is just my

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Answer from: litman876

$\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \infty$

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Limits

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

Limit Property [Multiplied Constant]: $\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$

Integrals

Definite IntegralsImproper Integrals

Integration Constant C

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

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx$

Step 2: Integrate

[Integrand] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \int\limits^{\infty}_{\frac{\pi}{2}} {7x^{\frac{-1}{2}}} \, dx$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = 7\int\limits^{\infty}_{\frac{\pi}{2}} {x^{\frac{-1}{2}}} \, dx$ [Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \lim_{b \to \infty} 7\int\limits^b_{\frac{\pi}{2}} {x^{\frac{-1}{2}}} \, dx$ [Integral] Reverse Power Rule: $\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \lim_{b \to \infty} 7(2x^{\frac{1}{2}}) \bigg| \limits^b_{\frac{\pi}{2}}$ [Limit] Rewrite [Limit Property - Multiplied Constant]: $\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = 7\lim_{b \to \infty} (2x^{\frac{1}{2}}) \bigg| \limits^b_{\frac{\pi}{2}}$ Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = 7\lim_{b \to \infty} (2b^{\frac{1}{2}} - \sqrt{2\pi})$ Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^{\infty}_{\frac{\pi}{2}} {\frac{7}{x^{\frac{1}{2}}}} \, dx = \infty$

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

Unit: Integration

Book: College Calculus 10e

Answer from: Quest

Do your own homework

Answer from: Quest

option b $7.5\ units$