IT304 - Integration

4. Definitions and Properties

Dr Robert Batzinger
Instructor Emeritus

2022-08-15

1 Comparison of Methods

1.1 Derivative - slope

\[\eqalign{\frac{dy}{dx} &=& \frac{d(4x - x^3)}{dx}\\ &=& 4-3x^2\\}\]

1.2 Key proof for derivative

\[\eqalign{ \frac{dy}{dx} &=& lim_{d\rightarrow 0} \left(\frac {4(x+d) - (x+d)^3 - (4x - x^3)}{d}\right)\\ &=&lim_{d\rightarrow 0} \left(\frac{4x + 4d -x^3 -3dx^2 - 3d^2x - d^3 - 4x + x^3}{d}\right)\\ &=&lim_{d\rightarrow 0} \left(\frac{ 4d -3dx^2 - 3d^2x - d^3}{d}\right)\\ &=&lim_{d\rightarrow 0} \left(4 -3x^2 - 3dx - d^2 \right) = 4 -3x^2 \\ }\]

1.3 Integration - area

\[\eqalign{ area_{rthnd} &=& h \left(f(x_1) + f(x_2) + ... + f(x_4)\right)\\ area_{midpt} &=& h \left({\Large f}\left(\frac{x_0 +x_1}{2}\right) + {\Large f}\left(\frac{x_1 +x_2}{2}\right) + ...+{\Large f}\left(\frac{x_3 +x_4}{2}\right)\right)\\ area_{trap} &=& \frac{h}{2}\left(f\left(x_0\right) + 2 f\left(x_1\right) + 2 f\left(x_2\right)+ + 2 f\left(x_3\right) + + 2 f\left(x_1\right)\right)\\ area_{simp}&=& \frac{h}{3}\left(f\left(x_0\right) + 4 f\left(x_1\right) + 2 f\left(x_2\right) + 4 f\left(x_3\right) + f\left(x_4\right)\right)\\ area_{int} &=& \int_{x_0}^{x_4} f(x)dx\\ }\]

1.4 Numerical Integration

Method Calculated Area
Right hand rule 3.50
Midpoint rule 4.125
Trapezoidal rule 3.75
Simpson rule 4
Integration by antiderivative 4

1.5 Application

  • Injection of dye in a vein to measure cardiac blood flow
  • Requires integrating dye concentration in arterial blood over time.
  • Also useful for measuring the flow of rivers and glacier flows.

\[\eqalign{\left(\frac{mg}{area}\right) &=&\frac{mg_{total}}{\left(\frac{mg}{Litre_{blood}}\right)\cdot sec}\\ &=& \left(\frac{mg}{sec}\right)\cdot\left(\frac{Litre_{blood}}{mg}\right)\\ &=& \frac{Litre_{blood}}{sec}}\]

1.6 Dye concentration

$ \[\begin{matrix} time & 5 & 7 & 9 & 11& 13\\ dye & 0 & 3.8 & 8.0 & 6.1 & 3.6\\ \\ time & 15 & 17 & 19 & 21 & 23\\ dye & 2.3 & 1.45 & 0.91 & 0.57 & 0.36 \\ \\ time & 25 & 27 & 29 & 31 & 33\\ dye & 0.23 & 0.14 & 0.09 & 0.05 & 0\\ \end{matrix}\]

$

1.7 Area of a rectangle

1.8 Calculations

  • Area by trapezoid Rule

\[\eqalign{\frac{h_0+h_1}{2dx} + \frac{h_1+h_2}{2dx} + \frac{h_2+h_3}{2dx}+ \frac{h_3+h_4}{2dx}&=& \frac{h_0 + 2h_1 + 2h_2 + 2h_3 + h_4}{2dx}\\ &=& \frac{6+12+12+12+6}{2} = \frac{48}{2} = 24\\}\]

  • Area by Simpson Rule

\[\eqalign{\frac{h_0+4h_1+h_2}{3dx} + \frac{h_2+h_3+h_4}{3dx}&=& \frac{h_0 + 4h_1 + 2h_2 + 4h_3 + h_4}{3dx}\\ &=& \frac{6+24+12+24+6}{3} = \frac{72}{3} = 24\\}\]

  • Area by antideriative

\[\int_0^{4} (6)dx=\displaystyle \bigg|_0^{4} 6x = 6\cdot 4 - 6\cdot 0 = 24 - 0 = 24\]

1.9 Integration of a line

1.10 Calculations

  • Area by trapezoid Rule

\[\eqalign{\frac{h_0+h_1}{2dx} + \frac{h_1+h_2}{2dx} + \frac{h_2+h_3}{2dx}+ \frac{h_3+h_4}{2dx}&=& \frac{h_0 + 2h_1 + 2h_2 + 2h_3 + h_4}{2dx}\\ &=& \frac{0+8+16+24+16}{2} = \frac{64}{2} = 32\\}\]

  • Area by Simpson Rule

\[\eqalign{\frac{h_0+4h_1+h_2}{3dx} + \frac{h_2+h_3+h_4}{3dx}&=& \frac{h_0 + 4h_1 + 2h_2 + 4h_3 + h_4}{3dx}\\ &=& \frac{0+16+16+48+16}{3} = \frac{96}{3} = 32\\}\]

  • Area by antiderivative

\[\int_0^{4} (4x)dx=\displaystyle \bigg|_0^{4} 2x^2 = 2\cdot 16 - 2\cdot 0 = 32 - 0 = 32\]

1.11 Area of Parabola

1.12 Calculations

  • Area by trapezoid Rule

\[\eqalign{\frac{h_0+h_1}{2dx} + \frac{h_1+h_2}{2dx} + \frac{h_2+h_3}{2dx}+ \frac{h_3+h_4}{2dx}&=& \frac{h_0 + 2h_1 + 2h_2 + 2h_3 + h_4}{2dx}\\ &=& \frac{0+6+8+6+0}{2} = \frac{20}{2} = 10\\}\]

  • Area by Simpson Rule

\[\eqalign{\frac{h_0+4h_1+h_2}{3dx} + \frac{h_2+h_3+h_4}{3dx}&=& \frac{h_0 + 4h_1 + 2h_2 + 4h_3 + h_4}{3dx}\\ &=& \frac{0+12+8+12+0}{3} = \frac{32}{3} = 10.6667\\}\]

  • Area by antiderivative

\[\int_0^{4} (4x-x^2)dx=\displaystyle \bigg|_0^{4} 2x^2-\frac{x^3}{3} = 2\cdot 16 -\frac{64}{3} - 2\cdot 0+\frac{0}{3} = 32 - 21.3333 = 10.6667\]

1.13 4th power

1.14 Calculations

  • Area by trapezoid Rule

\[\frac{h_0 + 2h_1 + 2h_2 + 2h_3 + h_4}{2dx} = \frac{0+6+8+12+0}{2} = \frac{26}{2} = 13\]

  • Area by Simpson Rule

\[\eqalign{\frac{h_0 + 4h_1 + 2h_2 + 4h_3 + h_4}{3dx} &=& \frac{0+12+8+24+0}{3} = \frac{44}{3} = 14.6667\\}\]

  • Area by antiderivative

\[\eqalign{\int_0^{4} (-\frac{x^4}{2}+\frac{7x^3}{2}-8x^2+8x)dx&=&\displaystyle \bigg|_0^{4} -\frac{x^5}{10}+\frac{7x^4}{8}-\frac{8x^3}{3}+4x^2\\ &=& -\frac{1024}{10} +\frac{7\cdot 256}{8} - \frac{8\cdot 64}{3}+4\cdot 16 = 14.93333\\}\]

1.15 Smaller intervals

1.16 Calculations

  • Area by trapezoid Rule

\[\frac{h_0 + 2h_1 + 2h_2 + 2h_3 + ... + h_8}{2dx} = \frac{0+2*(2.40625 + 3 + 3.28125 + 4 + 5.15625 + 6 + 5.03125) + 0}{2} = \frac{28.875}{2} = 14.4375\]

  • Area by Simpson Rule

\[\eqalign{\frac{h_0 + 4h_1 + 2h_2 + 4h_3 + ... + 4h_7+ h_8}{3dx} &=& \frac{0+ 4\cdot 2.40625 +2\cdot 3+4\cdot 3.28125 + 2\cdot 4 + 4\cdot 5.15625 + 2\dot 6 + 4\cdot 5.03125 + 0}{6}\\ &=& \frac{89.5}{6} = 14.91667\\}\]

  • Area by antiderivative

\[\eqalign{\int_0^{4} (-\frac{x^4}{2}+\frac{7x^3}{2}-8x^2+8x)dx&=&\displaystyle \bigg|_0^{4} -\frac{x^5}{10}+\frac{7x^4}{8}-\frac{8x^3}{3}+4x^2\\ &=& -\frac{1024}{10} +\frac{7\cdot 256}{8} - \frac{8\cdot 64}{3}+4\cdot 16 = 14.93333\\}\]

1.17 Even smaller intervals

1.18 Calculations

  • Area by trapezoid Rule

\[\frac{h_0 + 2h_1 + 2h_2 + ... + 2h_11 + h_12}{2dx} = \frac{44.1358}{3} = 14.71193\]

  • Area by Simpson Rule

\[\frac{h_0 + 4h_1 + 2h_2 + 4h_3 + ... + 4h_{11} + h_{12}{3dx} = \frac{134.3704}{9} = 14.93004\]

  • Area by antiderivative

\[\eqalign{\int_0^{4} (-\frac{x^4}{2}+\frac{7x^3}{2}-8x^2+8x)dx&=&\displaystyle \bigg|_0^{4} -\frac{x^5}{10}+\frac{7x^4}{8}-\frac{8x^3}{3}+4x^2\\ &=& -\frac{1024}{10} +\frac{7\cdot 256}{8} - \frac{8\cdot 64}{3}+4\cdot 16 = 14.93333\\}\]

1.19 Yet smaller intervals

1.20 Calculations

  • Area by trapezoid Rule

\[\frac{h_0 + 2h_1 + 2h_2 + ... + 2h_15 + h_16}{2dx} = \frac{118.4688}{4} = 14.80859\]

  • Area by Simpson Rule

\[\frac{h_0 + 4h_1 + 2h_2 + 4h_3 + ... + 4h_{15} + h_{16}}{3dx} = \frac{179.1875}{12} = 14.93229\]

  • Area by antiderivative

\[\eqalign{\int_0^{4} (-\frac{x^4}{2}+\frac{7x^3}{2}-8x^2+8x)dx&=&\displaystyle \bigg|_0^{4} -\frac{x^5}{10}+\frac{7x^4}{8}-\frac{8x^3}{3}+4x^2\\ &=& -\frac{1024}{10} +\frac{7\cdot 256}{8} - \frac{8\cdot 64}{3}+4\cdot 16 = 14.93333\\}\]

1.21 3/8 rule

1.22 END

1.23 End

PYU IT304 2022/1: Calculus ….. [3]