008c introduction to integral calculus 2.pdfx


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008c introduction to integral calculus 2.pdfx

  1. 1. Integral Calculus Consider a lamp. Integration gives an approximation of the volume of me base by cutting it to an infi)21'ra number of i7gt‘mize. j' thin slices. The symbol of integration: _l'. Thinl: of the integration symbol as an elongated S for “sum up. " top J‘ dL : L bottom Wheredl means alittle bit of the lamp — actually an infinite small piece. The equation justmeans, you sum up all the little pieces of the lamp from the bottom to the tap. the result is Z, the Volume of the whole lamp. Finding the. -Xrea under the Curve The most fimdamental meaning of integration is to add up, Andix-hen we depict integration on the grmh, we can see the adding up process as summing up of little bits of area to arrive at the total area under the curve. Finding the. -rea under the Cun'e The most fundamental meaning of integration is to add up. Andvchen we depict integration on the graph, we can see the adding up process as summing up of little bits of area to arrive at the total area under the curve. lhe shaded area can be calculated with the following integral: _| ': f ( X) dx. Look at the thin rectangle in the figure, it has a heightfixl and width of dx (3 little bit of . r), so the area (ierrgtiz times width) is given by f (_ X) - dx. The above integral tells you to add up tl1e areas of fl the narrow rectangular strips between a and 2: under the cur‘efiJc),
  2. 2. Ifboth 1 and _}' axis are both infauzt its an ar_ea If, on the other hand I-axis is in izsur: rt» and the y-axis is labeled in mile: per hour, then it is a distance. Or if the x-axis is labeled in / ‘raw: Kr» and the ‘Y-? ..‘Zl5 is in ,1*i. ’swi‘az: of electrical power, then the area under the curve is . i*i. ’su‘azr~i:9ur: of energy consumption between two points in time. Finding theVolume Inthe graph, the function . -1u". i_I gives the cross-sectional area of die ' ice of the Imp as the function of its heightmeasured from the bottom of the lamp. So this time, the ii-axis is labeled in ir: :.»': e: (that's the height from fllé bottom of the lamp), and the _‘. '-axis is labeled in square in: /3.3:. and thus each thin rectangle has a width measured in inches and height measured in square inches. Volume = (cross — sectional area)(_tl2z'rkr. es5) 15 V zf A(: h)dh o Which means that you addup the volumes of fl the thin slices from 0 to 15 inches (that is, from the bottom to the top of1an1p’s base) each slice haxing a volume given bj. ‘,(~.1_flz_J (its cross-sectional area) time a'F: [its height or thiclcness. )
  3. 3. Antidifferentiation: Difl'erentiation in Reverse . -ln11?a'_tfl‘+zrenti? atz'on is just differentiation back-wards. The derivative of 13 is 3x2, so the antiderivative of 3x2 is x’. Thelndefinitelntegnl The indefinite integral of a function_/ v'1gI, written as I f (x) dx, is me family of a1] antiderivatives of the function. For example, bme the derivative of 1:3 is 31:‘, the z'na'efin1'te integral of 3:2 is x3 + C, and we write: J. 312 dx = :3 + C The definite integral symbol contains two little numbers lie tell you to compute the area of the function between those two numbers, called the Ibnit: of integration The naked version of symbol f, indicates an indefinite integral or an antideriraIt’'e. TheConstant oflntegration The figure below shows the family of antiderivatives of 3x2, namely x3 + C. Note that th. is family of curves has an infinite number of curves. They go up and down forever and are infinitely dense. Indefininte integrals have the following properties which are easily verified by differentiation: (1) fdu= u+C (2) _[(du+dv— dw) : _|'du+ _| 'dv +_[dv. ' (3) fcduzcfdu
  4. 4. 'l'hePower Formula Consider the poweru’”1 where u is a continuous function of x and n is a constant other than -1. Differentiation: d(u”“) = (n + 1)u”du From whim we obtain: u”du = dwlm) n+1 Integrating, we obtain the general; power formula of integration: ufiél j'u"du= +C, n,= ‘-1 n+1 Example: Evaluate f(3x: — Gxm + %) dx. Solution: f(3x2 — 6:: +%) dx =3_fx2—6fx“’+9_[x" = [3-$1-[6-%l+[9-"-’ +c -a _ 3_ 3,"2_£ _x 41* x, +C We notice that whena function is integrated, the result is a function which involves an arbitrary constant called the constant of integration Ifsuffident conditions are the value of this constant can be determined and usedin the course of a solution. Reference: Calculus for Dummies by Mark Ryan 3' Wiley Pub, 2C 03 Elements of Calculus and Analytic Geometry by Francisco G. Reyes and Jenny L. Chua ti 1992 UST Publishing Houselklanila, Philippines.