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kipiarov [429]

Answer:

$\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Calculus

Derivatives

Derivative Notation

Solving Differentials - Integrals

Integration Constant C

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

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Step-by-step explanation:

*Note:

Ignore the Integration Constant C on the left hand side of the differential equation when integrating.

Step 1: Define

$\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}$

t = 1

s = 8

Step 2: Integrate

[Derivative] Rewrite [Leibniz's Notation]: $\displaystyle ds = (5t^3 + \frac{9}{t^2})dt$
[Equality Property] Integrate both sides: $\displaystyle \int {} \, ds = \int {(5t^3 + \frac{9}{t^2})} \, dt$
[Left Integral] Reverse Power Rule: $\displaystyle s = \int {(5t^3 + \frac{9}{t^2})} \, dt$
[Right Integral] Rewrite [Integration Property - Addition]: $\displaystyle s = \int {5t^3} \, dt + \int {\frac{9}{t^2}} \, dt$
[Right Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt$
[Right Integrals] Rewrite [Exponential Rule - Rewrite]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {t^{-2}} \, dt$
[Right Integrals] Reverse Power Rule: $\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{t^{-1}}{-1}) + C$
[Right Integrals] Rewrite [Exponential Rule - Rewrite]: $\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{1}{t}) + C$
Multiply: $\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} + C$