4.2 Fundamental Theorem of Calculus
Pre-Class 4.2A:
- Complete 4.1 Homework assignment: check and correct.
- Take notes on the videos and readings (use the space below).
- Work and check problems #1-2 in the 4.2 NOTES section.
- Complete the 4.2A Pre-Class Quiz.
Pre-Class 4.2B:
- Complete 4.2A Homework assignment: check and correct.
- Complete the 4.2B Pre-Class Quiz.
Introduction
From 4.1 homework problem #2:
Given the function $f(x)=4-0.16{{x}^{2}},$
approximate the area under the curve on the interval [0, 6] using 6 right rectangles.
Notes
Antiderivatives
General Antiderivative Formulas
$\begin{align}&\int u^n\;du\end{align}\;=\;\frac{u^{n+1}}{n+1}+C\;\;n\neq-1$
$\begin{align}&\int e^u\;du\end{align}\;=\;e^u+C$
$\begin{align}&\int\frac1u\;du\end{align}\;=\;\ln\left|u\right|+C$
$\begin{align}&\int E'\left[I(x)\right]I'(x)\;dx\end{align}=E\left[I(x)\right]+C$
$\begin{align}&\int\left[f(x)\right]^nf'(x)\;dx\end{align}=\frac{\left[f(x)\right]^{n+1}}{n+1}+C\;\;\;n\neq-1$
$\begin{align}&\int e^{f(x)}\cdot f'(x)\;dx\end{align}=e^{f(x)}+C$
$\begin{align}&\int{\frac{1}{f\left( x \right)}~{f}'\left( x \right)~dx~}\end{align}=~\ln \left[ f\left( x \right) \right]+C$
The Fundamental Theorem of Calculus
Let $f$ be continuous on [a, b]. If $F$ is any antiderivative for $f$ on [a,b], then $$\begin{align}&\int_a^bf\left(x\right)\operatorname dx\end{align}\;=F(b)-F(a).$$Evaluate the Definite Integral: Compare your answer to the area you calculated in 4.1 notes.
-
$S(x)=\begin{align}&\int_0^4x^2dx\end{align}$
$ = (\frac{{{x^3}}}{3} + C)│_0^4$
$=[\frac{{{4^3}}}{3} + C] - [\frac{{{0^3}}}{3} + C]$
$=21.33 + C - 0 - C$
$=21.33$
-
$S(x) =\begin{align}&\int_1^4 {2xdx}\end{align}$
$=(\frac{{2{x^2}}}{2} + C)│_1^4$
$= ({x^2} + C)|_1^4$
$=({4^2} + C) - ({1^2} + C)]$
$=16 + C - 1 - C$
$=15$
-
$S(x) = \begin{align}&\int_1^4 {{x^3}} dx\end{align}$
$=( \frac{{{x^4}}}{4} + C)│_1^4$
$=(\frac{{{4^4}}}{4} + C) - (\frac{{{1^4}}}{4} + C)$
$=64 + C - \frac{1}{4} - C$
$=63.75$
-
$S(x) = \begin{align}&\int_{0.5}^2 {\frac{1}{x}} \,dx\end{align} $
$=(ln \left| x \right| + C)│_{0.5}^2$
$= (\ln (2) + C) - (\ln (0.5) + C)$
$= \ln (2) + C - \ln (0.5) - C$
$= \ln (2) - \ln (0.5)$
$=\ln (\frac{2}{{0.5}}) = \ln (4)$
$\approx1.39$
-
$\begin{align}&\int_0^7 8x\;dx\end{align} $
$= (\frac{{8{x^2}}}{2}+C)│_0^7$
$= (4{x^2}+C)│_0^7$
$(4{\left( 7 \right)^2} + C) - (4{\left( 0 \right)^2} + C)$
$= 196 + C - 0 - C$
$= 196$
-
$\begin{align}&\int_2^3 4{x^3}\;dx\end{align}$
$= (\frac{{4{x^4}}}{4} + C)│_2^3$
$= {x^4} + C│_2^3$
$= ({3^4} + C) - ({2^4} + C)$
$= 81 + C - 16 - C$
$= 65$
-
$\begin{align}&\int_0^3 5{e^x}\;dx\end{align}$
$= (5{e^x}+C)│_0^3$
$= (5{e^3} + C) - (5{e^0} + C)$
$= 5{e^3} + C - 5(1) - C$
$= 5{e^3} - 5$
$= 5({e^3} - 1)$
$ \approx 95.43$
-
$\begin{align}&\int_1^{4\;} \frac{7}{x}\;dx\end{align}$
$= (7\ln| x|+C)│_1^4$
$= 7(\ln 4 + C) - 7(\ln 1 + C)$
$= 7\ln 4 + C - 7\ln 1 - C$
$= 7\ln 4 - 7\ln 1$
$\approx 9.70$
-
$\begin{align}&\int_3^7 \;\left( {2 - 4{x^2}} \right)\;dx\end{align}$
$= (2x - \frac{{4{x^3}}}{3} + C)│_3^7$
$= [2(7) - \frac{{4{{\left( 7 \right)}^3}}}{3} + C] - [2(3) - \frac{{4{{\left( 3 \right)}^3}}}{3} + C]$
$= (14 - 457.33 + C) - (6 - 36 + C)$
$= - 443.33 + 30$
$= - 413.33$
-
$\begin{align}&\int_4^{25} \frac{5}{{\sqrt x }}\;dx\end{align}$
$=\begin{align}&\int_{4}^{25}{5{{x}^{-1/2}}dx}\end{align}$
$= \left(\frac{{5{x^{1/2}}}}{({1/2})} + C \right)│_4^{25}$
$= (10{x^{1/2}}+C)|_4^{25}$
$= (10\sqrt {25} + C) - (10\sqrt 4 + C)$
$= 10(5) - 10(2)$
$= 50 - 20$
$= 30$
-
Cost: A company manufactures mountaineering 75-liter backpacks. The research department produced the marginal cost function $$B'\left( x \right) = 400 - \;\frac{x}{5}\quad\quad 0\; \le x\; \le 1000$$ where $B’(x)$ is in dollars and x is the number of backpacks produced per week. Compute the increase in cost when production level increases from 0 backpacks per week to 600 backpacks per week. Set up a definite integral and evaluate it.
$B=\begin{align}&\int_0^{600} {400 - \frac{x}{5}dx}\end{align}$
$B= (400x - \frac{{{x^2}}}{{5 \cdot 2}} + C)│_0^{600}$
$B=[400(600) - \frac{{{{600}^2}}}{{10}} + C] - [400(0) - \frac{{{0^2}}}{{10}} + C]$
$B= 240,000 - 36,000 + C - 0 - C$
$B= \$ 204,000$
The cost increased by \$204,000 when production increased from 0 to 600 backpacks per week.
-
Costs of Upkeep of a Marina: Maintenance costs for a marina generally increase as the structures at the marina age. The rate of increase in maintenance costs (in dollars per year) for a particular marina is given approximately by $$M'\left( x \right) = 30{x^2} + 2000$$ where x is the age of marina, in years, and M(x) is the total accumulated costs of maintenance for x years. Write a definite integral that gives the total maintenance costs from the third through the seventh year, and evaluate the integral.
$M=\begin{align}&\int_3^7 {30{x^2} + 2000 dx}\end{align}$
$M=(\frac{{30{x^3}}}{3} + 2000x + C)│_3^7$
$M= 10{x^3} + 2000x + C│_3^7$
$M= [10{\left( 7 \right)^3} + 2000(7) + C] - [10{\left( 3 \right)^3} + 2000(3) + C]$
$M= 3430 + 14000 + C - 270 - 6000 - C$
$M = \$ 11,160$
The total maintenance costs for the marina for years three through seven will be $\$11,160.$
