# Definite Integral

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```					         1'.
~i 5.2 The Definite Integral

2. L6    :=      t
t=l
1(:[,-1) 6.1:                [T,: =                ·1:,-1   is:1 len endpoinl and 6.:7: = 0.5]                                                  y

f(.\J=llIx-1
= 0.5 [1(1)              + I(J .5) + ./(2)
+ I(2.;i) + .1(:1) + /(;3·:"))1                           [.1(.1:) =            II1.r -    11                                        Il                                                        x

;:0;;   O.;i( -1 - ();JD45:1·l!) - IUIl1i8.S2H
- ll.08l7WJ:l               + OOC)Sli 12:1 + O.2527(j:.lO)

= 05( -1. n:U7:l17) ~ -O.RHiHG 1

The Ricmann sum represcnts the sum of the arcas of the two rectangles
abuve tl\(: .I:-axi, minus the SUIll of the areas of the four n:ctanglcs below
the ,·ax is; that is, the                    nl!! area         of thc rccwnglcs with respect to the .c-axis.

';
4. (a) R 6              L:
= ,=1 )'(1',) .6.1:                    [./.;    = :r,          is a right endpoint and 60                      = O.S]                                                  J1.\") C~   .\' -     ~     sin 2.1'
<I
= OS [./(O.S) t f(1)                         + .J (1.5) + I(2)
+ /('2'5) -I-               j'(:3»)      [J (t) =-=        J: -    2 sin2rJ
2

The RiclllRlln sum rerrcsents the slim of the nrC,JS of the four rcctangles

2.' 1 J.)          x
~bove the T-nxis minus thc sum of the arC,1S of the                                                 tW()      rect3ngles helow

the :r-axis.

8. (a) Using the ril'ht endpoints to approximate                                            .r~:l .r;(.r) dJ:, we have
(;

L: y(.I.,).6,
i=1
= l[y( -2) -+- g( -1)                          + 9(0) + 9(1) + g(2) + gel))

;:0;;   1 - 05 - 15 - 15 - 0.5                                  + 2.5   = -05
1(1) Using the leli endpoints to approximate                                            r~:19(1:) dr, we have
Ii

L
' -J
<;(:j'i- tl6J =            1[.£I ( -3)     + g( -2) + g( -1) + g(O) + g( 1) + 9(:2)1

;:0;;    2   +1 -        0.5 - 15 - 15 - 0.5 = -1
(c) Using the                   III idpoint     of each subi ntcrv31 lo appro>; imate                                J: 3.£1(:") d.c, we have
r,

L~
i::::.1 (/('7,) 6/                 = l[y( -25)           + q(--15) + q( -(5)                               +q(05)    + y(1.5) + q(2iJ)]

~ 1.5 -+- 0 - 1 - 175 - 1 -+- 0.5 = -1.75

15. 011 [0. r.. lim. .'~. .r, '-'.
I       L .1', sin "1"=}o·"·rsin:rd.1'.
"
1
L ~-                                    .I·J~ v /2J; + :[;2 ux.
,,-   ~rx;, 1=      1
17. On 1 ,0;,               n),""l>~ JillJ t
~~ 1      "'11
•      -.1
/ ••
(I: J2 6.1",     '-=

~             c',       .6 r -- /." -'_:"- iiI'.
16. On [I. ;11, ,,!iI~~ :;;": 1 -I-               .1:,   !.     --.     J       l    --.1'
18. On 10,2),                        lim
nlf\X"j, 1', -,(j
t
i '_I
(4 - :l(:r;)"   +- (j(r:)"j .."1.7:,     =0      ((4 _ 3.r2
, ()
+ 6.yO) dl;

```
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