# Chapter 3 Applications of Integration v21Mar2017(1).pdf - 3...

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1 The integrals we have been dealing with so far such as 212xdxxxc+=++or cossin2122x dxxc=+may be referred to as indefinite integrals. It is possible to actually evaluate integrals and come up with a numerical answer - these are called definite integrals[])a(F)b(F=)x(Fdx)x(fbaba-=Where F(x) is an antiderivative of f(x) (Note: There is no need to include the ‘+ c’ for a definite integral). So whereas 323x dxxc=+the definite integral 2[]3232xdxx3====322781933--Further examples: 31=)131()031(=]0cos31[)23cos(31=3cos313sin2020×--×---×--=πππxdxx
3 APPLICATION OF INTEGRATION 3.1 Definite Integrals
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2 724.12=)1e(32=32e32=e32e32=e32dxe233031010x3x3---=[]25)248()3927(xxxdx1x2x33223232=++-++=++++Exercise 3.1 Evaluate: 1.dx1x2x3231++2.dxx12213.dxxsin20π4.dxe2x3105.dx)7x2(x43221-6.π20cos xsin2 x7.+2132dx7xx38.dxex22x11-dx
3 3.2 AreasJust as the derivative of a function allows us to calculate its gradient at any point, the definite integral can be used to calculate the area under a graph. The area under the curve between x= a and x= b can be approximated by adding the areas of a number of “thin” rectangles of width δx.yδxf(x) a bThe area of each rectangle is its height multiplied by its width i.e. f(x) ×δIf we add the areas of all these rectangles together we get the expression Area xabfx=( ) δTo get a better approximation of the area we can make δx smaller and thus find the sum of the areas of a greater number of “thinner” rectangles. To find the exact area under the curve we take the limit as δx0 and find the sum of an infinite number of rectangles each with an infinitely small width. i.e. Area =limδxoxab=f(x) δthis expression is usually written as Area =baf(x) dx So the area under a curve can be found by use of the definite integral.
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4 Example: Find the area under the graph of y = x2from x= 0 to xNote: it is always helpful to sketch the graph: y = xy0 3 A = 0x2dx = 3033= 33- 0 = 9 Area = 9 square units. (Note that area is measured in square units)
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5 It is important to note that although the definite integral is used to find the area it does not always give you the area directly. For example: Find the area enclosed by the graph of y = x2- 4 and the xaxis. The definite integral -22x2- 4 dx = x32234--= 838838---+= --16316= -32= -1023The definite integral is equal to -1023. The area, however, is 1023square units ( we don’t speak of a negative area), the negative answer indicates the area is below the x -axis. (You may sketch the graph yourself to illustrate that this area is indeed below the x - axis In some cases we wish to find an area which is partly above and partly below the xaxis. Example: Find the area enclosed by the graph of y = sin x , 0 x1 2ππ23π2π-1 If we simply find the definite integral we get π20sin xdx= []-cosx02π= -cos 2π- -cos 0 = -1 + 1 = 0 The area is obviously not zero, but the definite integral is equal to zero. This is due to
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