111 Worksheets For Calculus 1 4-2
111 Worksheets For Calculus 1 4-2
111 Worksheets For Calculus 1 4-2
Arc Length.
The concepts of the method of Riemann Sum and integration are both directed at measuring total accumulation.
The length of a curve can be described as the total length of a combination of short curves. Any smooth curve that
p
is short enough would look like a line segment, and the length would be approximately (dx)2 + (dy)2 . Thus we
can get an arc length formula if we integrate the expression.
The Arc Length Formula for Graphs: Consider the graph y = f (x). If f 0 is continuous on the closed interval [a, b],
then the length of y = f (x), a x b, is
s
Sds =/ Cd✗THdñ =/ ¥j it .
dx Z b p Z b
dy
2
L= 1+ [f 0 (x)]2 dx = 1+ dx.
dx
=/Gyp -11 .
dy a a
A further application of the arc length formula lies in the arc length function defined by
Z xp
s(x) = 1 + [f 0 (t)]2 dt.
a
The length function is useful in finding out the exact point you would arrive at after travelling for a fixed distance.
dy
it
"
= ×
d-✗
11+1%4,5 =/ It hi
-
¥)
'
=
Itai
↳ Ji *+
ix. dx =
six * ×
-
'
i =
§ ¥ fit
-
=
%, *
1
Exercise 2. Find the exact length of the curve
y = ex , 0 x 1.
I "
loaf )
.
'
=
/ It e
"
Silke "
dx ≤
Jeanie secf-s.ec Tan f
'D
do
tanf
gtañle
)
Let e' =
,
.
do dd -
Sino
snoopy
=
¥
ei.dx-se.io do
Jeani¥ e
]
,
DO
d✗=sei0dO
=
siiowsno
tant
geantcel
I
a.
DO
% (1-050) cos't
E
dy =
Tx
SIX)=§ [ CHXJÉ (1+1)%1=1 -
d-
✗
f.
"
I" 4-1×7-3=3+252
↳
dx.fi#xdx
=
5
"
,
/ Hx -
deux)
✗
=
(23-+252)%-1 *
§ ( H✗)É
4
= .
=
IIIs IT ) -
=
13-1-5 -
% *
(b)
sci-fi / He d-l-ojli-it.li/I=FIc+xF3-lHDEf For a function to has inverse function
± -
£
inverse
51×1 ≤
Jux so when × ≥ -1 ( bijection ) *EJ¥f *, EEK
onto
sad is
strictly increasing when ×≥ -1
SAD is on -
to
-
one on ✗ 2-1
:
I ≤✗ ≤ 4 i.
six has inverse function
Average Value and Center of Mass.
Finding the average value of a function is a natural topic in many real world scenarios. It is important to remember
that a clear domain must be given before an average can be computed.
The center of mass of an object with a given density function is the average position of all point masses. We treat
the whole object as a system and we use the moment of the system about the origin to find the center of mass. For
a pipe-like object that covers the interval [a, b] of a given density ⇢(x), the mass equals
Z b
m= ⇢(x) dx
a
and the position of its center of mass equals
Z b
1
x= x⇢(x) dx.
m a
Exercise 4.
(a) Find the average value of f (x) = 25 x2 on the interval [0, 2].
(b) Find all values of c in the interval [0, 2] such that f (c) is equal to the average value.
[91
'
b- a Jab -51×1 DX =
2-10 [ 128 -
×
'
) DX -
÷ (25×-5×3) ? =
-1150 F)
- =
F- *
(b) 51×1=25 *-
=
¥ C-
4
} E- ±
¥
2
:c C- [0,2 ] :
[ =
-13 *
3
1
Exercise 5. Find the mass and the center of mass of a steel pipe with density function ⇢(x) = over the
p 1 + x2
interval [0, 3].
M
got '*d✗ tan"H) F
-
= = -
O -
[mass )
µ *
MI
=)? × .
*×
'
.
-
dxiz-lnkx.li?=i-lln4-ln1)=ln2
COM :
MI =
9¥ 3%2 .
Exercise 6. (Optional) Prove the Mean Value Theorem for Integrals : if f is continuous on [a, b], then there exists
c in [a, b] such that
Z b
1
f (x) dx = f (c).
b a a
Interpret this result geometrically.
* FTC I :
fix is cts
Lee Glx)
=)? 5- ctsde
Full -
Si -51-4 dt is dit
Then GCN is differentiable on [ a, b] for ✗ c- Iasb )
Applying MVT ,
7- C f I a. b) St .
Glb) -
Gla) =
G' (c) I b-a)
-514 ( b- a)
Jab said
-
b- a
=
-5 (c)
In
geometric sense , Sab -5 dx is the area bounded by the x-axis and Sid
514lb a)
rectangle with height
the of
is 914
length a)
-
area a
,
lb -