INTEGRATION (Reverse Process of Differentiation)



INTEGRATION (Reverse Process of Differentiation)Candidates should able to:Understand integration as the reverse process of differentiation, and integrate ax+bn (for any rational n except -1), together with constant multiples, sums and differences;Solve problems involving the evaluation of a constant of integration, e.g. to find the equation of the curve through (1, -2) for which dydx=2x+1;Evaluate definite integrals (including simple cases of ‘improper’ integrals, such as 01x-12dx and 1∞x-2dx);Use definite integration to find the area of a region bounded by a curve and lines parallel to the axes, or between two curves, and a volume of revolution about one of the axes.IntroductionIf we know the equation of a curve, say y=f(x) then by differentiation, we can find the gradient dydx. If instead, we are given function dydx, can we obtain the equation of the curve? This reverse process of differentiation (or process of anti-differentiation) is called INTEGRATION and we write:If dydx=x ? y=x dxSuppose dydx=2x. To find y, we require a function that when differentiated will give 2x. Clearly y=x2 is such a function, but there are many others, for example, y=x2+3, y=x2-1, y=x2+6, etc.We can say that y=x2+c, where c is a constant that gives 2x when differentiated. Thus if we integrate 2x, we obtain x2+c.Here are some simple cases for process of integration:dydx=2x ? y=x2+cdydx=4x3 ? dydx=x4 ?1.1 Basic rule of integrating algebraic functionGiven: dydx=axnThen, y=axn dxy=axn+1n+1+c provided n≠-1We integrate axn with respect to x. axn is the integrand and the result is called the integral. is the notation for integration.e.g. 1: Integrate the following w.r.t. x(a) 2x2(b) 1x3(c) x (d) 61.2 Integrating sums or differences of algebraic functionsThere are three important steps to take note:Change ALL terms with root sign to INDEX FORM;Denominators containing x term to be brought up to become numerator; andApply basic rule of integration (Section 1.1) to integrate term by term.e.g. 2: Find the integral of 1+x5+1x3-x. 1+x5+1x2-x dx=1+x5+x-2-x12 dx =1x+x5+15+1+x-2+1-2+1-x12+112+1+c =x+x66+x-1-1-x3232+c =x+x66-1x-23x32+ce.g. 3: Find an expression for y if dydx is given by:(a) 2x2-4x+3(b) 5x2-x(c) 2x-2x3-1 (d) 5-3x+4x41.3 Integrating simple products & quotients of functionsSteps to be done before starting to integrate:Express the simple products & quotients as sums or differences of functions; andNext integrate term by term applying the basic rule of integration to each term.e.g. 4: Integrate w.r.t. x.(a)x22x+1(b) x-12-x(c) 2-3x1+5x(d) x4-3x+12x3(e) 2x-1xdxExercise 1: Find(a) 12xdx(b) (x3+x)dx(c) x(x+1)dx(d) x+6(x-4)dx(e) 5x4dx(f) (10x4+8x3-6x2)dx(g) x4+1x2dx(h) (1-3x)xdx1.4 Integrating the POWER of a LINEAR function(ax+b)ndx=(ax+b)n+1a(n+1)+c. n≠-1, a≠0Take note: (ax+b) must be a linear function, if not, then the rule is not applicable. e.g. 5: Find (5x-1)2dx5x-12dx=5x-12+152+1+c =(5x-1)35(3)+c =(5x-1)315+ce.g. 6: Integrate the following w.r.t. x.(a) (x+3)-5(b) 2-5x6(c) 13-x2e.g. 7: Find an expression for y if dydx=5-2x .Exercise 2: Find(a) 10(4+3x)7dx(b) -12(3-2x)6dx(c) (3x+2)5dx(d) (2-5x)3dx(e) 2(3+4x)4dx(f) 6(1-2x)4dx1.5 Integration applied to coordinate geometrye.g. 8: Find the equation of the curve which passes through the point (2, 3) and for which dydx=3x2+x.e.g. 9: A curve passes through the point (1, 7) and the gradient of the curve at the point (x, y) is given by (5-2x). Find the equation of the curve. [Ans: y=5x-x2+3]Exercise 3: The gradient of a curve at the point (x,y) on the curve is given by 6x. If the curve passes through the point (1, 4), find the equation of the curve.Find the equation of the curve passing through the point (-2, 6) and having gradient function of 3x2-2.The gradient of a curve at the point (x,y) on the curve is given by 2(1-x) and the curve passes through the point (-1.5). Find the equation of the curve. The gradient function of a curve is given by (2x-3) and the curve cuts the x-axis at two points, A(5,0) and B. Find (i) the equation of the curve, and (ii) the coordinates of B.1.6 Problems involving 2nd order differentialse.g. 10: Find y as a function of x, given that d2ydx2=15x-2 and that when x=2, dydx=25 and y=20.e.g. 11: Find y as a function of x, given that d2ydx2=4-6x and that when x=2, dydx=-4 and y=7.Definite Integrals The abf(x)dx integral is known as the definite integral.For example, we have ddx2x3+3x=6x2+3.And so ? 23(6x2+3)dx=2x3+3x23 =233+33-[223+32] =63-22=41In general, we have: abf(x)dx=F(x)ba=Fb-F(a)Important note: For definite integral – No constant, c !!!e.g. 12: Evaluate the following definite integrals.(a)183xdx(b) 042x+1dx(c) 15x3-1x2dx(d) -24(x-2)2dxExercise 4: Evaluate the following definite integrals(a) 152xdx(b) -14(6-2x)dx(c) -13(3x-2)dx(d) -40(x2+x+1)dx(e) 491xdx(f) 14x4-x3+x-1x2dxApplication of Integration – Area Under a Curveyy=f(x) xx=bx=aThe area of the region enclosed by the curve y=f(x), the x-axis and the lines x=a to x=b is given byA=abydx or A=abf(x)dx,where f(x)≥0 for a≤x≤b.e.g. 13: Find the area of the shaded regions given below. y=4x5Area of A1==Area of A2==The sum of A1 and A2 gives the area bounded by the curve, the x-axis and the lines x=1 and x=3. ∴ 12ydx+23ydx=13ydxe.g. 14: Find the area of the shaded regions B1 and B2 in the following diagram. y=4x-x3B1B2Area of B1 =024x-x3dx=4x22-x4420=162-164-0=4By property of rotational symmetry, the area of B2 should be 4 units2.Note that -204x-x3dx=4x22-x440-2=0-162-164=-4This means that if the shaded region is below the x-axis, integration will give a negative value so we should only take the magnitude of the integral.Therefore, the total area of the shaded regions is not -224x-x3dx, which would give 4x22-x442-2= 162-164-162-164=0.e.g. 15: Find the area enclosed by the given curve, the x-axis, lines x=1 and x=3.(a)y=x2+2(b) y=3x2(c) A curve has the equation y=4x. Find the area of the region enclosed by the curve, the x-axis and the lines x=1 to x=6.y=4x. (d) A curve has the equation y=xx-1x-2. Find the total area of the shaded regions.3.1 Area between a Curve and the y-axisx=g(y) The area enclosed by the curve x=g(y) and the y-axis from y=a to y=b is given by A=abxdy or A=abg(y)dyfor x≥0 and a≤y≤b.e.g. 16(a): Find the shaded area in the diagram below.C1y=x2Area of C1 = 14ydy =e.g. 16(b): Find the area of C1 by finding the area of C2 using integration about the x-axis first.y=x2C2C1Area of C2 = ∴ Area of C1 = 3.2 Area between 2 curvesShaded area between 2 curves =abfx-g(x)dxNote that f(x) is the curve on top for a<x<b.If the graph of f(x) and g(x) is a straight line, the area of a triangle or trapezium can be used.y=x2e.g. 17: Find the area of the shaded regions.BA(2, 4)y=2xe.g. 18: Find the coordinate of A and the area of the shaded region.y=12x-3x2OAy=3xExercise 5: Application of Integration – AreaFind the shaded area of the following diagrams.y=x+3(a)41(b)y=x2-2x20(c)202y=4x-x32. (a)The diagram shows the shaded regions A and B enclosed by the curve x=y(2-y) and the y-axis. Find the area of each of the regions.2. (b) Calculate the area of the shaded region of the diagram below.3.. The diagram shows part of the curve y=x2-3x+6 and part of the line y=x+3. FindThe coordinates of the points A and B,The area of the region R.Volume of a RevolutionWhen the shaded region R enclosed by the curve y=f(x), the x-axis, the lines x=a and x=b is rotated completely about the x-axis, a solid is formed and its volume is called the volume of revolution and is given by V=πaby2dxe.g. 19: Find the volume of the solid formed when the region bounded by the curve y=x2+2, the lines x=1 and x=3 is rotated completely about the x-axis.y=x2+2e.g. 20: Show that the volume of the solid formed when the shaded region is rotated completely about the x-axis is 325π.y=42x+1Similarly, if a portion of the curve x=f(y) is rotated about the y-axis, volume of revolution will be V=πabx2dye.g. 21: The portion of the curve y=x2 between x=1 and x=2 is rotated through 360° about the x-axis and the y-axis respectively. Find the volumes created.(a)(b)4.1 Volume generated by the region bounded by 2 curves (or a line with a curve)e.g. 22: Refer to PYQ (Integration) Qn40.Exercise 6: Do Past Year Question (Integration) Qn23, 29Additional InformationWhen the area enclosed by the lines y=r, x=h, the x- and y-axes is rotated through 360° about the x-axis, the solid formed is a cylinder. yx=hy=rxOIts volume is π0hπ(r)2dx=πr20h1dx=πr2xh0=πr2hThe volume of a cone can also be obtained by finding the volume of the solid formed by rotating the area enclosed by the lines y=rhx, x=h and the x-axis.Improper IntegralsAn improper integral is one where the limits are unbounded e.g. a≤x≤∞ or 0≤x≤b. To find the shaded area enclosed by y=x-2, the x-axis and the line x=1, we find 1bx-2dx, then consider what happens when b becomes infinitely large. y=x-21bx-2dx=-1xb1=-1b-(-1)As b approaches ∞, 1b→0 and the value of the integral is convergent. ∴1∞x-2dx=1.To find the shaded area enclosed by y=x-2, the y-axis and the line x=1 where a is very, very small value.y=x-201x-2dx=lima→001x-2dx=lima→0-1x1a=-1--10→∞The integral is divergent and the area cannot be determined. ................
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