3 CCHHAAPTTEERR 1133 2 1 Definite Integrals - Math Help

1 337 C H A P T E R 1 3C H A P T E R 1 3 Definite Integrals Since integration can be used in a practical sense in many applications it is often useful to have Integrals evaluated for different values of the variable of integration. Frequently we wish to integrate an expression between some limits. The practical significance of this will be explained later in this chapter. For example, if we wish to integrate 22xx+ between 1x= and 3x= then we write 3212 xxdx+!which is 33213xx!"+#$%&. Mathematically this means find the value of 323xx+ when 3x= and then subtract the value of 323xx+ when 1x=. ()3332211122 99116333xxxdxx!"#$+=+=+%+=&'()*+, Similarly ()442333223 4272xdxx!!"#==!!=$%& And also []22211111 1 lnxdxdxxxxx+!

2 "=+=+#$%&'' ()()2ln21ln11ln2=+!+=+. In general ()()() bafxdxFbFa=!" where ()()'Fxfx=. 338 This is called the Fundamental Theorem of Calculus. Note that it is not necessary to include the constant of integration since the subtraction cancels out that constant. Such Integrals are called Definite Integrals because we are substituting Definite values of x. Worksheet 1 definite integrals 1. Evaluate the following Definite Integrals . You may use a calculator. a) 5212 1xdxx!" b) 80sin4 xdx!" c) 102 1dxx+! d) 1202 1xdxx+! e) 102 1dxx+! f) 1201 69dxxx++! g) 10 1xdxx+! h) 108 34dxx+! i) 108 34xdxx+! j) 108 34dxx+! k) 1220243 1xxdxxx++++! l) 10 1xxedxe+! m) 4cos0sin xexdx!"# n) 12301 xxdx!

3 " o) 1021 1exdxx!++" p) 0sin3 xdx!" q) ()3202 1dxx+! r) 241sin xdx!!+" 339 s) 224sincos xxdx!!" t) 2sin0cos xxedx!" u) 2420sec 1tanxdxx!+" v) 23202 1xxdxx++! w) 1220214 xxdx!" Answers to Worksheet 1 1. a) 4 b) c) d) e) f) 112 g) h) i) j) k) l) m) n) 29 o) p) 23 q) 112 r) s) t) u) !4 v) w) 16 340 Worksheet 2 1. Evaluate the following Integrals without using a calculator. a) 3224 1xdxx!" b) ()203sin xxdx!" c) 1201exdxx!+" d) 2312 exxdxx!" e) 4211 dxxx! f) 2ln21xedx+! 2. 23dyxdx= and 5y= when 2x=. Find y when 3x=. 3. 1dyxdx=+ and 12y= when 3x=. Find y when 4x=. 4. A particle starts with an initial velocity of 3 feet per second.

4 Its acceleration is ()31t+ feet/sec2, where t is the number of seconds from the start. Find the velocity after 2 seconds. 5. Evaluate the following Integrals without a calculator: a) 2log312 xdx! b) 102 1exdxx!++" c) ()12202 1xxdxx++! Answers to Worksheet 2 1. a) 144ln2+ b) 3 c) 12 d) 11e+ e) 712 f) 232e 2. 24 3. 5 4. 11 5. a) 1ln2 b) e c) 12 341 Applications of Definite Integrals to Area Example 1 Question: Find the area between the x-axis , the graph of 2yx= and 4x=. Answer: We are trying to find the area of !"#$. By elementary geometry: Area 481622 OAAB!!"#$%===. Note also that 44222002 4016xdxx!"==#=$%&. 342 Example 2 Question: Find the area bounded by 2yx=, 3x=, 7x=, and the x-axis.

5 Answer: We wish to find the area of 61444022 ADBCABCDAB++!"!"=#=#=$%$%&'&'. Now consider 772332 49940xdxx!"==#=$%& also. It appears as though area is related to the Definite integral. 343 Area Under a Curve as a Definite Integral Let ()fx be a positive continuous function as shown below. We will try to find area under the curve bounded by the y-axis , ()yfx= and the x-axis from 0 to x we are trying to find Area OAB!C. Let ()Ax represent the area. To evaluate ()Ax for different values of x it is helpful to investigate the derivative of ()Ax. By First Principles, the derivative of ()Ax ()()()0'limxAxxAxAxx!"+!#==! =lim!x"0 Area OAE!D#Area OAB!C!x =lim!x"0 Area CBE!D!x 344 Now consider the region CBE!

6 D. Remember that CDx=! is considered as a small change in x. If we think of CBE!D as a body of sand whose upper edge is BE! then it is clear that BE! can be smoothed out horizontally so that area of CBE!D=area of rectangle CFGD (see above). This is true also by the Intermediate Value Theorem. It is also clear that FG intersects BE! at some point P whose x co-ordinate lies between x and xx+!. Let P be (t,()ft) where xtxx!!+". Area CBE!D= Area CFGD=CD!(yco-ordinate of P )=!x"ft(). Substituting into on the previous page, it follows that ()()0'limxxftAxx!"#="()0limxft!"= In the limiting case as 0x!", tx!. ! ()()'Axfx=. It follows that ()() Axfxdx=!. 345 To find the area under a curve ()yfx=, bounded by xa=, xb= and the x-axis we note that this is represented by region S in the diagram: Area of region S is therefore () bafxdx!

7 Comment Students often think that ()0 afxdx! is the area of region R and ()0 bfxdx! is the area of region RS+ but strictly speaking this is not true. Area region ()0 aRfxdxk=+! for some constant k And similarly Area region ()0 bRSfxdx+=+!the same constant k It does however remain true that Area () baSfxdx=! since the constant k cancels out . 346 Example Find the area under 2yx= from 2x= to 4x= above the x-axis. Area 443222648562 1833333xxdx!"===#==$%&'( Example Find the area bounded by 211yx=+, 1x=!, 3x= and the x-axis find the area of shaded region R in the picture below: 347 Area =11+x2 dx!13"=Arctanx#$%& !13 ()Arctan3 Arctan1=!! 34!!"#=$$%&'( 712!=. To find the area of a region whose boundaries involve more complex curves it is helpful to consider a thin strip procedure as follows.))

8 Example Find the area bounded by 2yx= and 22yx=. Note that the intersection points O and A are (0,0) and (1,2) respectively. Draw a thin vertical strip in the region whose area is to be evaluated. 348 Consider the thin strip as though it were a rectangle whose area is ()()2 co-ordinate on 2 co-ordinate on 2 times yyxyyxdx!"=#=$% ()Yydx=! =2x!2x2()dx To evaluate the area of the region we need to add up an infinite number of thin strips whose width dx tends to the limit of 0. This is effected by 12022 xxdx!" because integration is a process of adding an infinite number of values. Note that the thin strips vary from 0x= to 1x= which are the limits of integration for the variable x. Note also that dx has changed its role from width of the strip to the variable of integration for reasons beyond the scope of this text.

9 Area of required region =2x!2x2()dx01" ()12302210033xx!"#$=%=%%%&'()*+,- 13= Note that in a later chapter a similar process will be used to evaluate volumes. 349 Sometimes it is more convenient (and even perhaps required) to draw thin horizontal strips to evaluate an area. Example Find the area (in the first quadrant) bounded by 1yx=, 2yx=, and 4y=. Note that point A, the intersection of 2yx= and 1yx= is (1,1). In this example it is better to draw a thin horizontal strip because using vertical strips would necessitate that the region be divided into two separate parts. The area of the thin horizontal strip is ()Xxdy! where X denotes the x co-ordinate of point P on 2yx= and x denotes the x co-ordinate of point Q on 1yx=.

10 350 Area of region is ()44111 Xxdyydyy!=!"" 43212162lnln4ln1333yy!"#$#$=%=%%%&'()()* +*+&',- != (approx.) Note also that the thin strips vary from 1y= to 4y= and hence are the limits of integration. It is sometimes assumed that area under a curve is simply obtained by integrating the function. Care however and understanding are required as illustrated by the following example. Example Consider the function ()()()322424644248fxxxxxx=!+=!!! as shown below. 351 Let s obtain the area 12AA+ the area on either side of the x-axis from 1x= to 3x=. If we evaluate 332142464 xxdx!+" we get ()()3431864812161921864xxx!"#+=#+##+$% ()()5757=! 0= It is clear that the area of region 12AA+ is not zero.