Monday, January 12, 2015
MAT 146
f (x)
x
1
cos(x)
-sin(x)
ex
ex
2x 3 + 5x
6x 2 + 5
3p
0
( 2x +1)
10 ( 2x +1)
5
( cos x ) (sin x )
Monday, January 12, 2015
ò f (x)dx
f '(x)
31
(sin x )
30
ß
x2
+C
2
sin(x) + C
4
ex + C
x 4 5x 2
+
+C
2
2
3p x
1
6
2x
+1
(
) +C
12
1
32
(sin x ) + C
32
é- (sin x ) 2 + 31cos2 x ù
MAT 146
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û
CALCULUS!
LIMITS
FUNCTIONS
PRE-CALCULUS!
Monday, January 12, 2015
MAT 146
Monday, January 12, 2015
MAT 146
Calculus II
MAT 146
Methods of Integration: Substitution
“When the integration process is not immediately obvious, it
may be possible to ‘reduce’ the integral to a well -known form
by means of substitution.”
In any attempt to evaluate an integral, we try to determine what function has
the integrand as its derivative. That is, f or ! f (x)dx , we seek a function F
such that F ! = f . In many cases we can identify F by inspection, because we
know a lot about functions and their derivatives.
We can look at ! x dx and quite readily determine that ! x dx =
2
2
1 3
x + C,
3
1 3
x + C satisfies the
3
requirement that F ! = f . Likewise, there is no calculation or symbol
manipulation required to know or determine that ! cos(x)dx = sin(x) + C,
because we know the derivative of sin(x) is cos(x).
because for f (x) = x 2 , the function F(x) =
Just as there are many cases for which we can readily identify an integral,
there are lots more integrals for which it is difficult to simply identify by
inspection the function that gives the integrand as its derivative. Hence the
need for methods of integration.
The method of substitution is widely used with a variety of integrals, but all
such integrals have one thing in common: Within the integrand appear both
a function and (some multiple of) its derivative. Consider the f ollowing
examples.
Monday, January 12, 2015
MAT 146
Monday, January 12, 2015
MAT 146
Monday, January 12, 2015
MAT 146