drill: find dy / dx
Post on 14-Jan-2016
41 Views
Preview:
DESCRIPTION
TRANSCRIPT
Drill: Find dy/dx
• y = x3 sin 2x
• y = e2x ln (3x + 1)
• y = tan-1 2x
• Product rule:• x3 (2cos 2x) + 3x2 sin (2x)• 2x3 cos 2x + 3x2 sin (2x)
• Product Rule• e2x (3/(3x +1)) + 2e2x ln (3x + 1)• 3e2x/(3x +1) + 2e2x ln (3x + 1)
22 41
2
21
2
xx
Antidifferentiation by Parts
Lesson 6.3
Objectives
• Students will be able to:– use integration by parts to evaluate indefinite and definite
integrals.– use rapid repeated integration or tabular method to evaluate
indefinite integrals.
Integration by Parts Formula
A way to integrate a product is to write it in the form
If u and v are differentiable function of x, then
.functionanother of aldifferentifunction one
. duvuvdvu
Example 1 Using Integration by Parts
Evaluate .cos dxxx xdvxu cos,Let ,dxdu xdxv cos
xv sin duvuvdvu
xdxxxxdxx sinsincos
Cxxx cossin
Example 1 Using Integration by Parts
Evaluate .5 3 dxxe x dxedvxu x3,5Let ,5dxdu dxev x3
xev 3
3
1
duvuvdvu
dxeexdxxe xxx 5
3
1
3
155 333
dxexe xx 33
3
15
3
5C
exe
xx
)
33(5
3
5 33
Ce
xex
x 9
5
3
5 33
Example 1 Using Integration by PartsEvaluate .15cos dxxx
dxxdvxu 15cos,Let
dxxvdxdu 15cos, 15sin5
1 x
duvuvdvu
dxxxxdxxx 15sin
5
115sin
5
115cos
Cxxx 15cos
5
1
5
115sin
5
1
Cxxx )15cos(25
115sin
5
1
Example 2 Repeated Use ofIntegration by Parts
Evaluate .2 dxex xdxedvxu x ,Let 2
dxevdxxdu x,2xev duvuvdvu
dxxeexdxex xxx 222
dxxeex xx 22
dxxeex xx 22
dxedvxu x ,Let dxevdxdu x, xev
)(22 dxexeex xxx
)(22 Cexeex xxx
Cexeex xxx 222
Example 2 Repeated Use ofIntegration by Parts
Evaluate .sin2 dxxx dxxdvxu sin,Let 2
dxxvdxxdu sin,2
xv cos duvuvdvu
dxxxxxdxxx 2coscossin 22
dxxxxx cos2cos2
xdxxxx cos2cos2
dxxdvxu cos,Let
dxxvdxdu cos, xv sin
)sinsin(2cos2 xdxxxxx
))cos(sin(2cos2 Cxxxxx
Cxxxxx cos2sin2cos2
Example 3 Solving an Initial Value Problem
• Solve the differential equation dy/dx = xlnx subject to the initial condition y = -1 when x = 1
.ln xdxx xdxdvxu ,lnLet
dxxvdxx
du ,1
2
2xv
It is typically better to let u = lnx
dxx
xxxdvu
1
2)
2(ln
22
dxxxx
2
1ln)
2(
2
Cx
xx
22
1ln)
2(
22
Cx
xx
4
ln)2
(22
C4
11ln)
2
1(1
22
C4
101
C4
3
4
3
4ln)
2(
22
xx
xy
DrillSolve the differential equation: dy/dx = x2e4x (This means you will need to find the anti-derivative of dy/dx = x2e4x )
.42 dxex xxedvxu 42 ,Let
dxevdxxdu x4,2
4
4xev
xdxee
xdvuxx
244
442
dxxeex x
x4
42
2
1
4
dxxeex x
x4
42
2
1
4
xedvxu 4,Let
dxevdxdu x4,
4
4xev
dx
eex
ex xxx
442
1
4
4442
C
exeex xxx
1642
1
4
4442
C
exeex xxx
3284
4442
Cexeex xxx
3284
4442
Example 4Solving for the unknown integral
xdxe x cos xdxdveu x cos,Let
dxxvdxedu x cos,
xv sin xdxexedvu xx sinsin
xdxexexe xxx sinsincosxdxdveu x sin,Let
dxxvedu x sin,xv cos
xdxexexexe xxxx coscossincos
xdxexexexe xxxx coscossincos
xexexexe xxxx coscossincos
xdxexexexe xxxx coscossincos
xexexe xxx cossincos2 C
xexexe
xxx
2
cossincos
Rapid Repeated Integration by PartsAKA: The Tabular Method
• Choose parts for u and dv.• Differentiate the u’s until you have 0.• Integrate the dv’s the same number of times.• Multiply down diagonals.• Alternate signs along the diagonals.
Example 5 Rapid Repeated Integration by Parts
Evaluate
u and its derivatives dv and its integrals
.4sin2 dxxx
2x x4sin
x2
2
0
4
4cos x
16
4sin x
64
4cos x
Cxx
xx
x
64
4cos2
16
4sin2
4
4cos2
Example 5 Rapid Repeated Integration by Parts
Evaluate .4sin2 dxxx Cxx
xx
x
64
4cos2
16
4sin2
4
4cos2
Cxxxxx 4cos32
14sin
8
14cos
4
1 2
Example 5 Rapid Repeated Integration by Parts
Evaluate
u and its derivatives dv and its integrals
.2 dxex x
2x xe
x2
2
0
xe
xe
xe
Ceexex xxx 222
Example 5 Antidifferentiating ln x
xdxln dxdvxu ,ln
xvdxx
du ,1
dxx
xxxxdx 1
lnln
dxxxxdx 1lnln
Cxxxxdx lnln
Example 6 Antidifferentiating sin-1 x
xdx1sin dxdvxu ,sin 1
xvdxx
du
,1
12
dxx
xxxxdx
2
11
1
1sinsin
dxx
xxx
2
1
1sin
dxx
xxx
2
1
1sin
dxx
duxdxduxu
2,2,1 2
x
du
u
xxx
2sin 1
uduxxduu
xx
2
111
2
1sin
1
2
1sin
CuxxCu
xx
2
11
2
1
1 sin
212
1sin
Cxxx 2
121 )1(sin
Homework
• Page 346/7: Day #1: 1-15 odd• Page 347: 17-24
top related