Drill: Find dy / dx. y = x 3 sin 2x y = e 2x ln (3x + 1) y = tan -1 2x Product rule: x 3 (2cos 2x) + 3x 2 sin (2x) 2x 3 cos 2x + 3x 2 sin (2x). Product Rule e 2x (3/(3x +1)) + 2e 2x ln (3x + 1) 3e 2x /(3x +1) + 2e 2x ln (3x + 1). Antidifferentiation by Parts. Lesson 6.3. - PowerPoint PPT Presentation
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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)
• 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.