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The concentration, C, in ng/ml, of a drug in the blood as a function of the time, t, in hours since the drug was administered is given by C = 15te−0.2t. The area under the concentration curve is a measure of the overall effect of the drug on the body, called the bioavailability. Find the bioavailability of the drug between t =0 and t =3

Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 45.713 \ \text{ng/mL}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

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]

U-Substitution

Integration by Parts:                                                                                               [tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]

  • [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Area of a Region Formula:                                                                                     [tex]\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx[/tex]

Step-by-step explanation:

*Note:

It is given that the area under the concentration curve is equal to the bioavailability.

Step 1: Define

Identify

[tex]\displaystyle C = 15te^{-0.2t} \\\left[ 0 ,\ 3 \right][/tex]

Step 2: Integrate Pt. 1

  1. Substitute in variables [Area of a Region Formula]:                                  [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt[/tex]
  2. [Integral] Rewrite [Integration Property - Multiplied Constant]:                [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 15 \int\limits^3_0 {te^{-0.2t}} \, dt[/tex]

Step 3: Integrate Pt. 2

Identify variables for integration by parts using LIPET.

  1. Set u:                                                                                                             [tex]\displaystyle u = t[/tex]
  2. [u] Differentiate [Basic Power Rule]:                                                             [tex]\displaystyle du = dt[/tex]
  3. Set dv:                                                                                                           [tex]\displaystyle dv = e^{-0.2t} \ dt[/tex]
  4. [dv] Exponential Integration [U-Substitution]:                                             [tex]\displaystyle v = \frac{-e^{-0.2t}}{0.2}[/tex]

Step 4: Integrate Pt. 3

  1. [Integral] Integration by Parts:                                                                       [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 15 \Bigg[ \frac{-te^{-0.2t}}{0.2} \bigg| \limits^3_0 - \int\limits^3_0 {\frac{-e^{-0.2t}}{0.2}} \, dt \Bigg][/tex]
  2. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 15 \Bigg[ \frac{-te^{-0.2t}}{0.2} \bigg| \limits^3_0 + \frac{1}{0.2} \int\limits^3_0 {e^{-0.2t}} \, dt \Bigg][/tex]

Step 5: Integrate Pt. 4

Identify variables for u-substitution.

  1. Set u:                                                                                                             [tex]\displaystyle u = -0.2t[/tex]
  2. [u] Differentiation [Basic Power Rule, Multiplied Constant]:                       [tex]\displaystyle du = -0.2 \ dt[/tex]
  3. [Limits] Switch:                                                                                               [tex]\displaystyle \left \{ {{t = 3 ,\ u = -0.2(3) = -0.6} \atop {t = 0 ,\ u = -0.2(0) = 0}} \right.[/tex]

Step 6: Integrate Pt. 5

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 15 \Bigg[ \frac{-te^{-0.2t}}{0.2} \bigg| \limits^3_0 - \frac{1}{0.04} \int\limits^3_0 {-0.2e^{-0.2t}} \, dt \Bigg][/tex]
  2. [Integral] U-Substitution:                                                                               [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 15 \Bigg[ \frac{-te^{-0.2t}}{0.2} \bigg| \limits^3_0 - \frac{1}{0.04} \int\limits^{-0.6}_0 {e^{u}} \, du \Bigg][/tex]
  3. [Integral] Exponential Integration:                                                               [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 15 \Bigg[ \frac{-te^{-0.2t}}{0.2} \bigg| \limits^3_0 - \frac{1}{0.04}(e^u) \bigg| \limits^{-0.6}_0 \Bigg][/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 15 \Bigg[ -8.23217 - \frac{1}{0.04}(-0.451188) \Bigg][/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle \int\limits^3_0 {15te^{-0.2t}} \, dt = 45.713[/tex]

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

Unit: Integration

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