4.2 Fundamental Theorem of Calculus
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.
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 $$\begin{align}&\int_a^bf\left(x\right)\operatorname dx\end{align}\;=F(b)-F(a).$$4.2 Video
Evaluate the Definite Integral.
-
$S(x)=\begin{align}&\int_0^4x^2dx\end{align}$
$ = (\frac{{{x^3}}}{3} + C)│_0^4$
$=[\frac{{{4^3}}}{3} + C] - [\frac{{{0^3}}}{3} + C]$
$=21.33 + C - 0 - C$
$=21.33$
-
$S(x) = \begin{align}&\int_1^4 {{x^3}} dx\end{align}$
$=( \frac{{{x^4}}}{4} + C)│_1^4$
$=(\frac{{{4^4}}}{4} + C) - (\frac{{{1^4}}}{4} + C)$
$=64 + C - \frac{1}{4} - C$
$=63.75$
-
$S(x) =\begin{align}&\int_1^4 {2xdx}\end{align}$
$=(\frac{{2{x^2}}}{2} + C)│_1^4$
$= ({x^2} + C)|_1^4$
$=({4^2} + C) - ({1^2} + C)]$
$=16 + C - 1 - C$
$=15$
-
$\begin{align}&\int_3^7 \;\left( {2 - 4{x^2}} \right)\;dx\end{align}$
$= (2x - \frac{{4{x^3}}}{3} + C)│_3^7$
$= [2(7) - \frac{{4{{\left( 7 \right)}^3}}}{3} + C] - [2(3) - \frac{{4{{\left( 3 \right)}^3}}}{3} + C]$
$= (14 - 457.33 + C) - (6 - 36 + C)$
$= - 443.33 + 30$
$= - 413.33$
-
$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$
-
$\begin{align}&\int_1^{4\;} \frac{7}{x}\;dx\end{align}$
$= (7\ln| x|+C)│_1^4$
$= 7(\ln 4 + C) - 7(\ln 1 + C)$
$= 7\ln 4 + C - 7\ln 1 - C$
$= 7\ln 4 - 7\ln 1$
$\approx 9.70$
-
$\begin{align}&\int_4^{25} \frac{5}{{\sqrt x }}\;dx\end{align}$
$=\begin{align}&\int_{4}^{25}{5{{x}^{-1/2}}dx}\end{align}$
$= \left(\frac{{5{x^{1/2}}}}{({1/2})} + C \right)│_4^{25}$
$= (10{x^{1/2}}+C)|_4^{25}$
$= (10\sqrt {25} + C) - (10\sqrt 4 + C)$
$= 10(5) - 10(2)$
$= 50 - 20$
$= 30$
-
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 $\$11,160.$
-
$\begin{align}&\int_{ - 1}^5 {(1 - 2x)dx}\end{align}$
$=(x -\frac{{2{x^2}}}{2} + C)|_{ - 1}^5$
$=(x - {x^2} + C)|_{ - 1}^5$
$=(5 - 25 + C) - ( - 1 - {\left( { - 1} \right)^2} + C)$
$= (- 20 + C) - ( - 1 - 1 + C)$
$= - 20 + C + 2 - C$
$= - 18$
-
$\begin{align}&\int_0^3 {({x^3} + 2{x^2} - {e^x})dx}\end{align}$
$=(\frac{{{x^4}}}{4} + \frac{{2{x^3}}}{3} - {e^x} + C)|_0^3$
$= \left( {\frac{{{3^4}}}{4} + \frac{{2{{\left( 3 \right)}^3}}}{3} - {e^3} + C} \right) - \left( {\frac{{{0^4}}}{4} + \frac{{2{{\left( 0 \right)}^3}}}{3} - {e^0} + C} \right)$
$= \left( {20.25 + 18 - {e^3} + C} \right) - \left( { - 1 + C} \right)$
$= \left(18.164 \right) - (-1)$
$= 19.164$
-
$\begin{align}&\int_1^3 {\frac{1}{x}dx}\end{align}$
$=( \ln \left| x \right| + C)|_1^3$
$= (\ln 3 + C) - (\ln 1 + C)$
$= \ln 3 + C - \ln 1 - C$
$= \ln 3 - \ln 1$
$\approx 1.1$
-
$\begin{align}&\int_0^1\left(\sqrt x+\sqrt[3]{x^2}\right)\operatorname dx\end{align}$
$=\begin{align}&\int_0^1\left(x^{1/2}+x^{2/3}\right)\;\operatorname dx\end{align}$
$=( \frac{x^{3/2}}{3/2} + \frac{{{x^{5/3}}}}{{5/3}} + C)|_0^1$
$=(\frac{2}{3}{x^{3/2}} + \frac{3}{5}{x^{5/3}} + C)|_0^1$
$= \left[ {\frac{2}{3}{{\left( 1 \right)}^{3/2}} + \frac{3}{5}{{\left( 1 \right)}^{5/3}} + C} \right] - \left[ {\frac{2}{3}{{\left( 0 \right)}^{3/2}} + \frac{3}{5}{{\left( 0 \right)}^{5/3}} + C} \right]$
$= \frac{2}{3} + \frac{3}{5} + C - C$
$= \frac{{19}}{15} \approx 1.27$
-
$\begin{align}&\int_0^9 {(3\sqrt x + 2x + 1)dx}\end{align}$
$= \begin{align}&\int_0^9 {(3{x^{1/2}} + 2x + 1)dx}\end{align}$
$= (\frac{{3{x^{3/2}}}}{{3/2}} + \frac{{2{x^2}}}{2} + 1x + C)|_0^9$
$= (2{x^{3/2}} + {x^2} + x + C)|_0^9$
$= [2{\left( 9 \right)^{3/2}} + {9^2} + 9 + C] - [2{\left( 0 \right)^{3/2}} + {0^2} + 0 + C]$
$= [54 + 81 + 9 + C] - [C]$
$= 144 + C - C$
$= 144$
-
$\begin{align}&\int_1^e {\frac{2}{x}} dx\end{align}$
$= (2\ln \left| x \right| + C)|_1^e$
$= (2\ln e + C) - (2\ln 1 + C)$
$= (2\ln e + C) - (2(0) + C)$
$= 2\ln e + C - 0 - C$
$=2\ln e$
$= 2(1) = 2$
-
$\begin{align}&\int_1^3\left(\frac{4x^3-8x^2+2x-3}x\right)\operatorname dx\end{align}$
$=\begin{align}&\int_1^3\left(4x^2-8x+2-\frac3x\right)\;\operatorname dx\end{align}$
$=(\frac{{4{x^3}}}{3} - \frac{{8{x^2}}}{2} + 2x - 3\ln \left| x \right|+C)|_1^3$
$= [\frac{{4{{\left( 3 \right)}^3}}}{3} - 4{\left( 3 \right)^2} + 2(3) - 3\ln3+C] - [\frac{{4{{\left( 1 \right)}^3}}}{3} - 4{\left( 1 \right)^2} + 2(1) - 3\ln 1+C]$
$= (36 - 36 + 6 - 3\ln 3+C) - (\frac{4}{3} - 4 + 2 - 0+C)$
$= 2.70 +C - ( - 0.67+C)$
$= 2.70+C + 0.67-C$
$= 3.37$
-
$\begin{align}&\int_2^8\left(5x^{1/5}-2x^{2/3}-2x^{-2}+\frac7x-6\right)\operatorname dx\end{align}$
$=(\frac{{5{x^{6/5}}}}{{6/5}} - \frac{{2{x^{5/3}}}}{{5/3}} - \frac{{2{x^{ - 1}}}}{{ - 1}} + 7\ln \left| x \right| - 6x+C)|_2^8$
$=(\frac{{25}}{6}{x^{6/5}} - \frac{6}{5}{x^{5/3}} + 2{x^{ - 1}} + 7\ln \left| x \right| - 6x+C)|_2^8$
$=\left[ {\frac{{25}}{6}{{\left( 8 \right)}^{6/5}} - \frac{6}{5}{{\left( 8 \right)}^{5/3}} + \frac{2}{8} + 7\ln 8 - 6(8)+C} \right] - \left[ {\frac{{25}}{6}{{\left( 2 \right)}^{6/5}} - \frac{6}{5}{{\left( 2 \right)}^{5/3}} + \frac{2}{2} + 7\ln 2 - 6(2)+C} \right]$
$= - 21.07+C - ( - 0.39+C)$
$= - 21.07+C + 0.39-C$
$= - 20.68$
-
The rate of change of annual U. S. factory sales of electronics from 1990 through 1996 can be modeled by the equation $$s\left( t \right) = - 0.23{t^3} + 2.257{t^2} - 1.51t + 42.8$$ in billions of dollars per year where t is the number of years since 1990. Evaluate $\begin{align}&\int_0^6 s\left( t \right)\end{align}$ and interpret your answer.
$\begin{align}&\int_0^6 {( - 0.23{t^3} + 2.257{t^2} - 1.51t + 42.8)dt}\end{align}$
$=(\frac{{ - 0.23{t^4}}}{4} + \frac{{2.257{t^3}}}{3} - \frac{{1.51{t^2}}}{2} + 42.8t + C)|_0^6$
$= (- 0.0575( 6 )^4 + 0.752333( 6)^3 - 0.755( 6 )^2 + 42.8(6) + C) - (0 + C)$
$= 317.604 + C - C$
$= 317.604$
U.S. factory sales of electronics from 1990 through 1996 were \$317.604 billion.
-
The rate of change of the length of the average hospital stay between 1980 and 1996 can be modeled by the equations below where t is the number of years since 1980. Determine the value of the following definite integrals and interpret your answers. Note: $s(t)$ is in days per year.
$s(t)=\begin{cases} 0.028t-0.23 & 0\leq t\leq 10 \\ -0.0408t+0.30883 & 10\lt t\leq 16 \\ \end{cases}$
- $\begin{align}&S(t)=\int_0^{10}s(t)\;dt\end{align}$
$=\begin{align}&\int_0^{10} {(0.028t - 0.23)dt}\end{align}$
$= (\frac{{0.028{t^2}}}{2} - 0.23t + C)|_0^{10}$
$=[\frac{{0.028{{\left( {10} \right)}^2}}}{2} - 0.23(10) + C] - [0 - 0 + C]$
$= - 0.9 + C - C$
$= - 0.9$
The length of the average hospital stay decreased by 0.9 days from 1980-1990.
- $\begin{align}&\int_{10}^{16}s(t)\;dt\end{align}$
$=\begin{align}&\int_{10}^{16} {(-0.0408t+0.30883)dt}\end{align}$
$= (\frac{{-0.0408{t^2}}}{2} + 0.30883t + C)|_{10}^{16}$
$=(\left.-0.0204t^2+0.30883t+C\right)|_{10}^{16}$
$= [ - 0.0204{\left( {16} \right)^2} + 0.30883(16) + C] - [ - 0.0204{\left( {10} \right)^2} + 0.30883(10) + C]$
$= - 0.28112 + C - 1.0483 - C$
$ = - 1.33$
The length of the average hospital stay decreased by 1.33 days from 1990 to 1996.
c. $\begin{align}&\int_0^{16} s\left( t \right)\;dt\end{align}$From the answers to a & b: $\quad - 0.9 + - 1.33 = - 2.23$
The length of the average hospital stay decreased by 2.23 days from 1980 to 1996.
- $\begin{align}&S(t)=\int_0^{10}s(t)\;dt\end{align}$
-
$\begin{align}&\int_{ - 1}^2 {(3{x^2} - 2x + 5)\;dx}\end{align}$
$=(\frac{{3{x^3}}}{3} - \frac{{2{x^2}}}{2} + 5x + C)|_{ - 1}^2$
$=( {x^3} - {x^2} + 5x + C)|_{ - 1}^2$
$= ({2^3} - {2^2} + 5(2) + C) - ({\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} + 5( - 1) + C)$
$= (8 - 4 + 10 + C) - ( - 1 - 1 - 5 + C)$
$= 14 + C + 7 - C$
$= 21$
-
$\begin{align}&\int_0^4 {(3{e^x} - {x^2} + 3\;)dx}\end{align}$
$= (3{e^x} - \frac{{{x^3}}}{3} + 3x + C)|_0^4$
$= [3{e^4} - \frac{{{4^3}}}{3} + 3(4) + C] - [3{e^0} - 0 + 0 + C]$
$= (3{e^4} - \frac{{64}}{3} + 12 + C) - (3(1) + C)$
$= 3{e^4} - \frac{{28}}{3} + C - 3 - C$
$= 3{e^4} + \frac{{ - 37}}{3}$
$\approx 151.46$
-
$\begin{align}&\int_2^6 {\frac{{ - 4}}{x}\;dx}\end{align}$
$= ( - 4\ln \left| x \right| + C)|_2^6$
$= ( - 4\ln 6 + C) - ( - 4\ln 2 + C)$
$= - 4\ln 6 + C + 4\ln 2 - C$
$= - 4\ln 6 + 4\ln 2$
$\approx - 4.39$
-
$\begin{align}&\int_{ - 1}^1 {(\sqrt[3]{x} + \sqrt[5]{{{x^2}}})\;dx}\end{align}$
$\begin{align}&\int_{ - 1}^1( {{x^{1/3} + {x^{2/5}}} } )\;dx\end{align}$
$= \left(\frac{{{x^{4/3}}}}{{4/3}} + \frac{{{x^{7/5}}}}{{7/5}} + C\right)|_{ - 1}^1$
$=\left( \frac{3}{4}{x^{4/3}} + \frac{5}{7}{x^{7/5}} + C\right)|_{ - 1}^1$
$= \left(\frac{3}{4} + \frac{5}{7} + C\right) - \left(\frac{3}{4} - \frac{5}{7} + C\right)$
$= \frac{{10}}{7} \approx 1.43$
-
$\begin{align}&\int_{ - 1}^8 {(3\sqrt[3]{x} + x - 3)\;dx}\end{align}$
$\begin{align}&\int_{ - 1}^8 {(3{x^{1/3}} + x - 3)\;dx}\end{align}$
$= \left(\frac{{3{x^{4/3}}}}{{4/3}} + \frac{{{x^2}}}{2} - 3x + C\right)|_{ - 1}^8$
$= \left(\frac{9}{4}{x^{4/3}} + \frac{{{x^2}}}{2} - 3x + C\right)|_{ - 1}^8$
$= (36 + 32 - 24 + C) - (\frac{9}{4} + \frac{1}{2} + 3 + C)$
$= 44 + C - 5.75 - C$
$= 38.25$
-
$\begin{align}&\int_1^{2e} {\frac{{ - 5}}{x}\;dx}\end{align}$
$= (- 5\ln \left| x \right| + C)|_1^{2e}$
$= ( - 5\ln 2e + C) - ( - 5\ln 1 + C)$
$= (- 5\ln 2e + C) - (0 + C)$
$= - 5\ln 2e + C - 0 - C$
$= - 5\ln 2e$
$= - 8.466$
-
$\begin{align}&\int_1^4\left(\frac{4x^3-8x^2+2x-3}{2x^2}\right)\operatorname dx\end{align}$
$\begin{align}&\int_1^4 \left(2x - 4 + \frac{1}{x} - \frac{3}{2}{x^{ - 2}}\right)\;dx\end{align}$
$\left({x^2} - 4x + \ln \left| x \right| + \frac{3}{{2x}}+C\right)|_1^4$
$\left[ {{{\left( 4 \right)}^2} - 4(4) + \ln 4 + \frac{3}{8}}+C \right] - \left[ {{{\left( 1 \right)}^2} - 4(1) + \ln 1 + \frac{3}{{2(1)}}}+C \right]$
$\left( {16 - 16 + \ln 4 + \frac{3}{8}}+C \right) - \left( {1 - 4 + 0 + \frac{3}{2}}+C \right)$
$\approx (1.76+C )- ( - 1.5+C)$
$\approx 1.76+C + 1.5-C$
$\approx 3.26$
-
$\begin{align}&\int_1^4 {(\frac{5}{{{x^3}}} - \frac{2}{x} + \frac{3}{{\sqrt x }} + x + 1)\;dx}\end{align}$
$\begin{align}&\int_1^4 {(5{x^{ - 3}} - \frac{2}{x} + 3{x^{-1/2}} + x + 1)\;dx}\end{align}$
$= \left(\frac{{5{x^{ - 2}}}}{{ - 2}} - 2\ln \left| x \right| + \frac{{3{x^{1/2}}}}{{1/2}} + \frac{{{x^2}}}{2} + x + C\right)|_1^4$
$= \left(\frac{{ - 5}}{{2{x^2}}} - 2\ln \left| x \right| + 6\sqrt x + \frac{{{x^2}}}{2} + x + C\right)|_1^4$
$= \left( {\frac{{ - 5}}{{2(16)}} - 2\ln 4 + 6\sqrt 4 + \frac{{16}}{2} + 4 + C} \right) - \left( {\frac{{ - 5}}{2} + 0 + 6 + \frac{1}{2} + 1 + C} \right)$
$\approx (23.84375 - 2\ln 4 + C) - (5 + C)$
$\approx 21.071 + C - 5 - C$
$\approx 16.071$
-
Square Pharmaceuticals Limited, the flagship company of Square Group, is a pharmaceutical company in Bangladesh. Taka is the Bangladeshi currency (\$1=77.88 taka).
Monthly revenue is given by the function $R(t)=400-20t^\frac12.$
Monthly cost is given by the function $C(t)=40-40t^\frac12.$
Find the change in net profit for Square Pharmaceuticals from January 2011 through December 2013.
Source: https://www.academia.edu/7825243/Application_of_Mathematics_in_Real_Life_Business$P(t) = R(t) - C(t)$
$P(t) = (400 - 20{t^{1/2}}) - (40 - 40{t^{1/2}})$
$\begin{align}&\int_0^{36} \left(360 +20{t^{1/2}}\right)\;dt\end{align}$
$=\left(360t + \frac{{20{t^{3/2}}}}{{3/2}} + C\right)|_0^{36}$
$=\left(360t +\frac{40}{3}{t^{3/2}} + C\right)|_0^{36}$
$=15840 - 0$
The total change in net profit for 2011-2013 is $15,840$ taka.
4.2 Lecture
4.2 Group Work
4.2 Additional Practice
Use the Fundamental Theorem of Calculus to evaluate each of the following integrals.