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Kalkulus II (1 - 2)
1.
Kalkulus II Teguh
Budi P, M.Si Sesion #01-02 JurusanFisika FakultasMatematikadanIlmuPengetahuanAlam
2.
Outline The basic integral formula
Integration by parts 1/10/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 2
3.
Integration Techniques
© 2010 Universitas Negeri Jakarta | www.unj.ac.id | 3 1/10/2011
4.
When we find
the area under a curve by adding rectangles, the answer is called a Rieman sum. The width of a rectangle is called a subinterval. The entire interval is called the partition. subinterval partition Subintervals do not all have to be the same size. 1/10/2011 4 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
5.
If the partition
is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by . As gets smaller, the approximation for the area gets better. subinterval partition if P is a partition of the interval 1/10/2011 5 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
6.
is called the
definite integral of over . If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by: 1/10/2011 6 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
7.
Leibnitz introduced a
simpler notation for the definite integral: Note that the very small change in x becomes dx. 1/10/2011 7 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
8.
It is called
a dummy variable because the answer does not depend on the variable chosen. upper limit of integration Integration Symbol integrand variable of integration (dummy variable) lower limit of integration 1/10/2011 8 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
9.
Integration by parts
u = x dv= exdx du = dx v = ex 1/10/2011 9 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
10.
Integration by parts
u = lnxdv= x2dx du = 1/x dx v = x3 /3 1/10/2011 10 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
11.
v = x Integration by parts u = arcsin x dv= dx 1/10/2011 11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
12.
Integration by parts
u = x2 dv = sin x dx du = 2x dx v = -cos x u = 2xdv = cos x dx du = 2dx v = sin x 1/10/2011 12 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
13.
Thank You 1/10/2011
© 2010 Universitas Negeri Jakarta | www.unj.ac.id | 13