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Midterm

MAT136H1 Study Guide - Midterm Guide: Trigonometric Substitution, Fo Tan, Partial Fraction Decomposition


Department
Mathematics
Course Code
MAT136H1
Professor
Anthony Lam
Study Guide
Midterm

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Midterm Study Guide for MAT136
Antiderivatives
Definition: Any function F such that
F'
(
x
)
=f(x)
is called an antiderivative of f
E.g.,
dy
dx sinx=cos x
So
cos x
is the derivative of
sin x
. Hence,
sin x
is the antiderivative of
cos x
.
Note that
sin x+1
is another antiderivative of
cos x
, since
dy
dx (sinx+1)=cos x
.
Similarly,
(sinx +2)
is another antiderivative of
cos x
, since
dy
dx (sinx+2)=cos x
.
In general,
(where C may be any constant) is an antiderivative of
cos x
since
dy
dx (sinx+C)=cos x
.
Remark: if
F(x)
is an antiderivative of
f
(
x
)
,
then
F
(
x
)
+C
(where C is an
arbitrary constant) is also an antiderivative of
f(x)
. This
F
(
x
)
+C
is called the most
general antiderivative of
f(x)
, or the indefinite integral of
f(x)
, and is denoted by
f(x)dx
. The
f(x)
appearing in
f(x)dx
is called the integrand.
Note:
f(x)dx=F
(
x
)
+C
says the same thing as
F(x)=f(x)
dy
dx ¿
.
f(x)dx=F
(
x
)
+C
is called the integration formula. To prove that this is true, just
show that
F(x)=f(x)
dy
dx ¿
.

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

Note: If
f
(
x
)
dx=F
(
x
)
+C
, then we’ll say “The definite integral of
f
(
x
)
is
F
(
x
)
+C
” or “The most general antiderivative of
f
(
x
)
is
F
(
x
)
+C
” or “When we
integrate
f
(
x
)
with respect to
x
, we’ll get
F
(
x
)
+C ”.
Some Standard Integration Formulas:
1.
0dx=C
2.
k dx =k
(
x
)
+C
(for any constant k)
3.
xrdx=xr+1
r+1+C
(for any constant
r ≠1
)
4.
1
xdx=ln
|
x
|
+C
5.
exdx=ex+C
6.
ln¿a¿+C
axdx=ax
¿
(for any constant
a>0, a ≠ 1
)
7.
sin x dx=cos x+C
8.
cos x dx=sin x+C
9.
sec2x dx=tan x+C
10.
csc2x dx =cot x+C
11.
sec x tan x dx=sec x +C
12.
csc x cot x dx=csc x+C
13.
1
1+x2dx=arctan x+C
14.
1
(
1x2
)
dx=arcsin x+C
Theorem:
1.
1
a2+x2dx=1
aarctan
(
x
a
)
+C
(where a is a positive constant)

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

2.
1
a2x2dx=1
aarcsin
(
x
a
)
+C
(where a is a positive constant)
Note: these are special cases of 13 and 14 (above), respectively. To prove them, simply
derive the right side.
Theorem:
1.
(
f
(
x
)
+g
(
x
)
)
dx=¿f
(
x
)
dx+g
(
x
)
dx
¿
2.
(
f
(
x
)
g
(
x
)
)
dx=¿f
(
x
)
dxg
(
x
)
dx
¿
3.
kf
(
x
)
dx=kf
(
x
)
dx
(for any constant k)
E.g.,
x2+x3dx=x2dx+x3dx=x3
3+x4
4+C
E.g.,
5x2dx=5x2dx=5
(
x3
3
)
+C=5x3
3+C
Hence:
f
(
x
)
dx=F
(
x
)
+C
E.g., Find the indefinite integral of
f
(
x
)
=
(
x+2
) (
x1
)
x
Solution: The indefinite integral of
f
(
x
)
=f
(
x
)
dx=
(
x+2
) (
x1
)
xdx
¿x2+x2
xdx
¿x+12
xdx=x2
x+x2 ln
|
x
|
+C
E.g., If
f' '
(
x
)
=12 x212 x , f '
(
2
)
=11
and
f
(
1
)
=2
, find f(x).
Solution:
f'
(
x
)
=f''
(
x
)
dx=
(
12 x212 x
)
dx=12
(
x3
3
)
12
(
x2
2
)
+C .
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