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====322781933Further 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.dxx1221∫3.dxxsin20∫π4.dxe2x310∫5.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 δx→0 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 2ππ23π2π1 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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