Integration With Initial Conditions

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\CancelColor }[0]{} \newcommand {\cancelto }[0]{1} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$ $% This file was *autogenerated* from digInIntegrationWithIntitalConditions.sagetex.sage with % sagetex.py version 2015/08/26 v3.0-92d9f7a %de9efd6b7c4856ca17bac31f886d1bad% md5sum of corresponding .sage file (minus "goboom","current_tex_line",and pause/unpause lines)$

Suppose we know that $f"$ is the derivative of some function $f$ , is there a way to explicitly find $f$ ? The answer is yes, provided that we have some additional information:

$\int f' \d x = F(x)+C$

Where $F(x)$ is some antiderivative of $f'(x)$ and $C$ is an arbitrary constant. The problem is that $F(x)$ and $f(x)$ are not necessarily the same, just that they are both antiderivatives of $f'$ . What we do know is they differ by a constant. For convenience, write

$F(x)+C_{0}=f(x)$

where $C_{0}$ is a specific constant. In particular:

$C_{0} = F(x)-f(x)$

If we know a particular value of $f(x)$ then we can find $C_{0}$ . This particular value that we know is called an initial condition.

Suppose that $f'(x)=3x^7$ and $f(1)= 2$ , find $f(x)$ .

First we will find the indefinite integral of $f'$ : $\begin{align*} \int 8 x^7 \d x &= 3 \int x^7 \d x\\ &= 3 \cdot \answer[given]{\frac{x^8}{8}}+C. \end{align*}$

So we now know that $f(x)=3 \cdot \answer [given]{\frac {x^8}{8}}+C.$ Now we use the initial condition, $f(1)=2$ to find the specific value of $C$ $2=f(1)=3 \cdot \answer [given]{\frac {1^8}{8}}+C$ So $C=2-\frac {3}{8} = \answer [given]{\frac {13}{8}}$ Therefore $f=\answer [given]{\frac {3x^{8}}{8}+\frac {13}{8}}$

Marginal Revenue and Marginal Cost

Let’s apply what we’ve learned about integrating with initial conditions to working with marginal revenue and marginal cost.

Suppose that $R(q)$ is a total revenue function, then as money is earned only from selling products, $R(0)=0$ Likewise, if $C(q)$ is a total cost function, then $C(0)=C_{0}$ where $C_{0}$ is the fixed cost.

In both of these cases, we have a standard initial condition that we may use.

Suppose that $R'(q)=500+3q^{2}$ is a marginal revenue function. Find the total revenue function $R(q)$ .

First we will find the indefinite integral of $R'$ : $\begin{align*} \int R'(q) \d q &= \int 500+3q^{2} \d q\\ &= \answer[given]{500q+q^{3}+C} \end{align*}$

So we now know that $R(q)=500q+q^{3}+C$ Now we use the initial condition, $R(0)=)$ to find the specific value of $C$ $0=R(0)=0+C$ So $C= \answer [given]{0}$ Therefore $R(q)=500q+q^{3}$

Suppose that $C'(q)=200+q$ is a marginal cost function. Find the total cost function $C(q)$ if fixed costs are 1000.

First we will find the indefinite integral of $C'$ : $\begin{align*} \int C'(q) \d q &= \int 200+q \d q\\ &= \answer[given]{200q+\frac{q^{2}}{2}+C} \end{align*}$

So we now know that $R(q)=200q+\frac {q^{2}}{2}+C$ Now we use the initial condition. Since the fixed costs are 1000, this means $C(0)=1000$ . We can now use this to fnd the specific value of $C$ $1000=C(0)=0+C$ So $C= \answer [given]{1000}$ Therefore $C(q)=200q+\frac {q^{2}}{2}+1000$