Download Lecture Notes for Section 6.1

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Matrix calculus wikipedia , lookup

Sobolev space wikipedia , lookup

Function of several real variables wikipedia , lookup

Riemann integral wikipedia , lookup

Automatic differentiation wikipedia , lookup

Generalizations of the derivative wikipedia , lookup

History of calculus wikipedia , lookup

Itô calculus wikipedia , lookup

Clenshaw–Curtis quadrature wikipedia , lookup

Lebesgue integration wikipedia , lookup

Chain rule wikipedia , lookup

Multiple integral wikipedia , lookup

Integral wikipedia , lookup

Fundamental theorem of calculus wikipedia , lookup

Transcript
Calc 2 Lecture Notes
Section 6.1
Page 1 of 4
Section 6.1: Review of Integration Formulas and Techniques
Big idea: With some creative algebra, you can do a lot of “new-looking” integrals by
manipulating the integrand to match integral formulas from Calculus 1.
Big skill: You should be able to manipulate the integrands of the integrals in this section using
algebra and substitution so that they look like the integral formulas from Calculus 1.
Review of Integral Formulas from Calculus 1:
1
 x dx  r  1 x  c for r ≠ -1
 sin xdx   cos x  c
 sec xdx  tan x  c
 sec x tan xdx  sec x  c
 e dx  e  c
r
r 1
2
x
x
1
 1 x

2
dx  tan 1 x  c   cot 1 x  c
1
x 1
2
x

x
1
x 1
1
2
dx  sec 1 x  c
dx  cosh 1 x  c
dx  sech 1 x  c
1 x
 cosh xdx  sinh x  c
2
1
 x dx  ln x  c for x ≠ 0
 cos xdx  sin x  c
 csc xdx   cot x  c
 csc x cot xdx   csc x  c
 tan  x  dx  ln sec  x   c
2


1
1 x
1
2
1 x
2
1
 1 x
2
dx  sin 1 x  c   cos 1 x  c
dx  sinh 1 x  c
dx  tanh 1 x  c
 sinh xdx  cosh x  c
 sech
2
xdx  tanh x  c
And don’t forget integration by substitution (the chain rule in reverse):
d
F  x  f  x
dx
Now let’s look at what happens when we take the derivative of F composed with another
function u:
du  x 
d
F u  x   f u  x  
 f u  x    u  x 
dx
dx
So we see that F  u  x   is the antiderivative of f  u  x    u  x  , which can be written as
If f(x) is a function with an antiderivative F(x), then
 f  x   dx  F  x   c
and
 u  x   f u  x   dx  F u  x    c .
Thus, when we see an integrand that is the product of a composition of functions and the
derivative of the inner function, we’ll know that its antiderivative is simply the antiderivative of
the outer function of the composition (still composed with the inner function).
Calc 2 Lecture Notes
Section 6.1
Page 2 of 4
The key skill in this section is to figure out what formula an integral is most similar to, and then
use algebra (and substitution) to make the integrand look like the formula.
Examples:
1. To evaluate   2 x3  7  dx , expand the integrand so that you can use
3
the power rule formula:
1
 x dx  r  1 x
r
r 1
c.
1
dx , do some algebra and then a substitution on the integrand so that
 x2
1
du  tan 1 u  c .
you can use this formula: 
2
1 u
2. To evaluate
a
2
Calc 2 Lecture Notes
Section 6.1
Page 3 of 4
1
3. To evaluate

4. To evaluate
x
2
x
2
dx , do some algebra and then a substitution on the integrand
5  6 x  x 2
1
du  sin 1 u  c .
so that you can use: 
2
1 u
x 1
dx , simply use the (Calculus 1) technique of integration by
 2x  4
substitution.
5. To evaluate
1
 1 u
2
x3
dx , break the integral apart, and manipulate it so you can use
 2x  4
du  tan 1 u  c on the second piece.
Calc 2 Lecture Notes
6. Evaluate

ln  x 
2x
dx .
3
7. Evaluate  e 2ln  x  dx .
1
8. Evaluate  sin 2  x  dx .
Section 6.1
Page 4 of 4