Mathematics, 14.02.2020 19:33 JarretteWhitaker

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

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Answer from: kahnerp6448

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

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

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

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals

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

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

U-Substitution

Integration by Parts: $\displaystyle \int {u} \, dv = uv - \int {v} \, du$

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

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

Step-by-step explanation:

*Note:

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

Step 1: Define

Identify

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

Step 2: Integrate Pt. 1

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

Step 3: Integrate Pt. 2

Identify variables for integration by parts using LIPET.

Set u: $\displaystyle u = t$ [u] Differentiate [Basic Power Rule]: $\displaystyle du = dt$ Set dv: $\displaystyle dv = e^{-0.2t} \ dt$ [dv] Exponential Integration [U-Substitution]: $\displaystyle v = \frac{-e^{-0.2t}}{0.2}$

Step 4: Integrate Pt. 3

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

Step 5: Integrate Pt. 4

Identify variables for u-substitution.

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

Step 6: Integrate Pt. 5

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

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

Unit: Integration

Answer from: Quest

hi,

6x times 4 is 24x and 5 times 4 is 20, so the answer is 24x+20.

hope this and for a little humor i am groot.