Linear Functions



Area under the CurveTo approximate the area between a function f(x) and the x-axis, create rectangles that best fit and find the sum of the area of these rectangles. A curve above the x-axis is said to enclose “positive” area while a curve below the x-axis encloses “negative” area.Riemann Sum: LRAM, MRAM, RRAMRiemann Sum uses regular intervals when creating rectangles to approximate the area under the curve, Rectangular Approximation Method (RAM). LRAM creates rectangles with the height determined by the upper left-hand corner intersecting with the function. In MRAM, the function intersects at the upper middle part of the rectangle. In RRAM, the height of each rectangle is determined by the upper right-hand corner. LRAM, MRAM, and RRAM can be seen in that respective order below.Sn=k=1nf(ck)Δxn=number of rectanglesΔx=width of rectangles fck=height of each rectnalgeTrapezoidal RuleTrapezoidal rule uses trapezoids to approximate the area between a function f(x) and the x-axis. The height, h, of each trapezoid is the constant interval Δx and the bases are the y values at each interval.T=h2(y0+2y1+2y2+…+2yn-1+yn)Definite IntegralAs the width of each rectangle Δx decreases, the accuracy of the area approximation increases. Creating more rectangles over an interval [a, b] yields a better answer. Thus, breaking an interval into infinitely lim?n→∞ many rectangles yields an integral from a to b. dx keeps track of variable of integration and comparable to ?x.abfxdx=limn→∞k=1nf(ck)Δx?Indefinite IntegralsAntidifferentiation is the inverse of differentiation: an antiderivative of f(x) is any function F(x) whose derivative is equal to f(x): F'x=f(x). There are many functions with the same derivative; these functions usually differ by a constant C. Therefore, the family of antiderivatives of f(x) are Fx+C. Thus, the indefinite integral of f(x):fxdx=Fx+C3x2+4x-8dx=x3+2x2-8x+Ccscxcotxdx=-cscx+C1x2+1dx=tan-1x+Cdxx=lnx+CFundamental Theorem of CalculusPart 1If f is continuous on [a, b], then the functionFx=axftdtHas a derivative at every point x in [a, b], and F'x=ddxaxftdt=f(x)Part 2 (Integral Evaluation Theorem)If f is continuous at every point of [a, b], and if F is an antiderivative of f on [a, b], thenabfxdx=Fb-F(a)Integral Propertiesf(x) and g(x) are separate functionsfx±gxdx=fxdx±gxdxConstant Multiple: k is any real number kfxdx=kfxdxAdditivity: for any b, such that a<b<cabfxdx+bcfxdx=acfxdxOrder of Integrationabfxdx=-bafxdxZero Integralaafxdx=0Domination: f(x)≥g(x)abfxdx≥abgxdxBasic Integralsk dx=kx+Cxndx=xn+1n+1+C if n ≠-11xdx=x-1dx=lnx+Cexdx=ex+Caxdx=axlna+Clnxdx=xlnx-x+Clogaxdx=xlnx-xlna+Ctanxdx=-lncosx+Ccotxdx=lnsinx+Csinxdx=-cosx+Ccosxdx=sinx+Csec2xdx=tanx+Csecxtanxdx=secx+Ccsc2xdx=-cotx+Ccscxcotxdx=-cscx+Cdxx2+1=tan-1x+Cdx1-x2=sin-1x+Cdxxx2-1=sec-1x+Cu-SubstitutionThis is the chain rule for integrals and it is used for composite functions f(g(x)) and products. Let u=g(x), then fgxg'xdx= f(u)duor, with limits,abfgxg'xdx= g(a)g(b)f(u)duExample:x2x3+8dxu=x3+8 du=3x3dx ? du3=x2dxx2x3+8dx=13udu=13?23u32+C=29(x3+8)3+CPiecewise Integrationhx=f(x)x≥bg(x)x<bIf a<b<c, then the integral of h(x) from a to c isachxdx=abfxdx+bcgxdxAbsolute Value IntegrationRewrite as piecewise function. Solve for x inside the absolute value symbol. The inside of the absolute value should be positive with x≥ and negative with x<. Then, integrate as a piecewise function.f(x)=f(x)x≥b-f(x)x<bIf a<b<c, then the integral of f(x) from a to c isacfxdx=abfxdx-bcfxdxOriginal EquationGiven a derivative and an initial value, to find the original equation: Take the antiderivative.Substitute the initial value.Solve for the constant, C.Rewrite equation.ExamplesFind the particular solution to the equation dydx=ex-6x2 whose graph passes through the point (1, 0).y=ex-6x2dx=ex-2x3+CApply the initial condition (1, 0).0=e1-2(1)3+CSolve for C.C=2-eRewrite Equation.y=ex-2x3+2-eFind the particular solution to the equation dydx=2x-sec2x whose graph passes through the point (0, 3).y=2x-sec2xdx=x2-tanx+CApply the initial condition (0, 3).3=(0)2-tan(0)+CSolve for C.C=3Rewrite Equation.y=x2-tanx+3Physical MotionDisplacement: stVelocity: vt=s'tAcceleration: at=v't=s''(t)Conversely,vt=v0+0taxdxst=s0+0tvxdxTotal distance traveled from t=a to t=b isabv(t)dtExampleA particle travels with velocity vt=t-2sint m/sec for 0≤t≤4 sec. What is the particle’s displacement? What is the total distance traveled?displacement=04t-2sintdt≈ -1.45 metersdistance traveled=04t-2sintdt≈ 1.91 metersConsumption over TimeLet C(t) be a rate at which something in consumed over a period from t=a to t=b. Then, the total consumption can be modeled by the integralTc=abCtdtHooke’s LawHooke’s Law for springs says that the force it takes to stretch or compress a spring x units from its natural (unstressed) length is a constant times x: F=kx. The amount of Work needed to stretch or compress a spring x units from its natural (unstressed) length can be found by solving the integral below.0bFxdx=0bkxdxArea13335043815If f and g are continuous with f(x)≥g(x) throughout [a, b], then the area between the curves from a to b is the integral of [f-g] from a to b, A=abf(x)-g(x)dx3038475203200ExampleFind the area of the region enclosed by the parabola y=2-x2 and the line y=-x. The limit of integration are found by finding where the two equations intersect 2-x2=-x?x=-1, 2.A=-122-x2-(-x)dx=92Thus, the area of the region is 4.5 square units.Volume463551905If a solid in space is oriented so that the area of a cross-section perpendicular to the x-axis is given by A(x). Then, the volume of solid bounded by the planes x=a and x=b isabA(x)dxDisk Method463555080Volume of a solid that revolves around the x-axis bounded by x=a and x=b isabπradius2dx=πab(fx)2dxShell Method1339853810Volume of a solid that revolves around the y-axis bounded by x=a and x=b isab2πshellradiusshellheightdx=2πabxf(x)dxWasher Method368046075565-889075565If f(x)≥g(x) between a and b, the volume of a solid that revolves around the x-axis bounded by x=a and x=b isabπouterradius2-πinnerradius2dx=πab(fx)2-(gx)2dx3147695-44450Mean (Average) Value Theorem for IntegralsIf f is continuous on [a, b], then at some point c in [a, b],fc=1b-aabfxdxThe Mean Value Theorem for Integrals is the average value of a function over an interval. This is not to be confused by the Mean Value Theorem for Derivatives. The MVT for Derivatives is the average rate of change of a function over an interval. ExampleWhat is the average value of the cosine function on the interval [1, 5]?fc=15-115cosxdx=-0.450The average value of cosine is –0.450.If the average value of the function f on the interval a, b is 10, then abfxdx?10=1b-aabfxdxabfxdx is equal to 10(b-a).Separable Differential EquationsA differential equation expressed in the form dydx=f(x)g(y) is called separable, if it is possible to separate the variables completely. To integrate separable differential, separate the variables and rewrite in the form1g(y)dy=f(x)dxThen integrate each side separately. It is sufficient to write only one constant, C.Exampledydx=6x2y2?dyy2=6x2dxdyy2=6x2dx-1y=2x3+Cy=-12x3+Cdydx=y+22x-3?dyy+2=2x-3dxdyy+2=(2x-3)dxln(y+2)=x2-3x+Cy+2=Cex2-3xy=Cex2-3x-2Exponential Growth and DecayLaw of Exponential ChangeIf y changes at a rate proportional to the amount present dydt=ky and y0 is the initial amount at t=0, then y=y0ekt. The constant k is the growth constant if k>0 or the decay constant if k<0.Half-LifeThe half-life, h, of a radioactive substance with rate constant k is h=ln2k.Modeling Growth with other Bases y=y0bht where the base b is the multiplied by y0 over the time period 1h.ExampleScientists who use carbon-14 dating use 5700 years for its half-life. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed.y=y0bht?0.9y0=y012t5700?t=5700ln0.9ln0.5=866The sample is about 866 years old.Slope FieldsA slope field graphically shows the family of solution to a differentiable equation in dydx=f(x)g(y) form. To construct a slope field, for each point (x,y), draw a short line segment with slow dydx. To interpret a slope field (or find a particular solution) at a given point, follow the shape of the slope field through that point.A slope field for the differential equation dydx=x+yThe same slope field with the graph of the solution through (2,0) superimposed. ................
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