Page 179 - Calculus Workbook For Dummies

163
Chapter 9: Getting into Integration

Q. Estimate the area under f x = x + 3 x from 5. Replace the general function expres-
2
^ h
0 to 5 using 20 right rectangles. Use sigma sion with your specific function,
2
^ h
notation where appropriate. Then use f x = x + 3 x.
sigma notation to express the area approxi- 1 20 1 2 1
! c >
mation when you use n rectangles. 4 i 1 4 i + cm 3 4 imH
=
A. For 20 rectangles: ≈ 84.2; for n rectangles: 6. Simplify, pulling everything to the out-
side, except functions of i.
2
475 n + 600 n + 125 2
6 n 2 1 20 1 1 20 1
= !c i + ! cm 3 im
=
1. Sketch the function and the first few 4 i 1 4 4 i 1 4
=
20 20
1
1
and the last right rectangles. See the = ! 1 i + ! 3 i
2
following figure. 4 i 1 16 4 i 1 4
=
=
20
1
= ! i + 3 20 i !
2
y 64 i 1 16 i 1
=
=
2
f(x) = x + 3x
7. Compute the area, using the following
60
50 rules for summing consecutive integers
40 and consecutive squares of integers.
30
20 The sum of the first n integers equals
^
10 n n + 1h
x 2 , and the sum of the squares
-10 1 2 3 4 5 6
of the first n integers equals
n n + 1 2 ^h n + 1h
^
.
2. Add up the area of 20 rectangles. Each 6
has an area of base times height. So for So now you’ve got:
starters you’ve got
1 2 20 +
^
^
1 f 20 20 + h ^ $ 1h p + 3 f 20 20 + 1h p
$
! _ base heighti 64 6 16 2
20 rectangles
$
$
1 20 21 41 3
$
3. Plug in the base and height information = 64 c 6 m + 16 $ 10 21
to get your sigma summation. 17220 630
^ 5 - 0h 1 = 384 + 16
The base of each rectangle is , or .
20 4 . 84 .2
1
So you’ve got ! 1 height= ! height.
20 4 4 20 8. Express the sum of n rectangles instead
1 of 20 rectangles.
The height of the first rectangle is f c m,
4 1
2 3 Look back at Step 5. The outside and
the second is f c m, the third is f c m, 1 4
4 4 the two s inside come from the width
and so on until the last rectangle, which 4
has a height of f 5 ^ h. This is where the of the rectangles that you got by dividing
index, i, comes in. You can see that the 5 (the span) by 20. So the width of each
1 rectangle could have been written as 5 .
jump amount from term to term is , 20
1 4
so the argument will contain a i: To add n rectangles instead of 20, just
1 1 4 5
f ! c im. replace the 20 with an n — that’s . So
n
4 20 4
5
the three 1 s become . At the same
4. Create the sum range. 4 n
time, replace the 20 on top of the !
i has to equal 1 to make the first term with an n:
1
f c m. And because you’ve got to add n 2
4 5 5 5
up 20 rectangles, i has to run through n ! c > n i + cm 3 n imH
i 1=
20 numbers, so it goes from 1 to 20:
1 20 1
f ! c im.
4 i 1= 4
continued