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This presentation is at a Undergraduate in Science (Math, Physics, Engineering) level.
Please send comments and suggestions to solo.hermelin@gmail.com, thanks! For more presentations, please visit my website at http://www.solohermelin.com
Solving boundary value problems using the Galerkin's method. This is a weighted residual method, studied as an introduction to the Finite Element Method.
This is a part of a series on Advanced Numerical Methods.
Solving second order ordinary differential equations (boundary value problems) using the Least Squares Technique. Contains one numerical examples from Shah, Eldho, Desai
Mathematics and History of Complex VariablesSolo Hermelin
Mathematics of complex variables, plus history.
This presentation is at a Undergraduate in Science (Math, Physics, Engineering) level.
Please send comments and suggestions to solo.hermelin@gmail.com, thanks! For more presentations, please visit my website at http://www.solohermelin.com
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Integration is a part of Calculus.
This is just a short presentation on Integration.
It may help you out to complete your academic presentation.
Thank You
* Recognize characteristics of parabolas.
* Understand how the graph of a parabola is related to its quadratic function.
* Determine a quadratic function’s minimum or maximum value.
* Solve problems involving a quadratic function’s minimum or maximum value.
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Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
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Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
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Teknik-Pengintegralan
1.
2. Ingat Aturan Rantai pada Turunan :
Jika kedua ruas diintegralkan, maka diperoleh
)('))(('))(( xgxgfxgf
dx
d
=
dxxgxgfdxxgf
dx
d
)('))(('))(( ∫∫ =
dxxgxgfCxgf )('))(('))(( ∫=+
dari definisi integral tak tentu
3. Misal u = g(x), maka du = g’(x)dx
Disubstitusi ke atas diperoleh
Cxgfdxxgxgf +=∫ ))(()('))(('
Cufduuf +=∫ )()('
4. 1. Mulai dengan fungsi yang diintegralkan
2. Kita misalkan u = g(x)
3. Hitung du
4. Substitusi u dan du
5. Integralkan
6. Ganti u dengan g(x)
5. Hitunglah
Jawab
Misalkan u = 3x + 5 , maka du = 3 dx , dx = 1/3 du
Substitusi ke fungsi di atas diperoleh
dxx )53sin( +∫
CxCuududxx ++−=+−==+ ∫∫ )53cos(cossin)53sin(
6. Hitunglah
Jawab
Misalkan u = -3x2
+ 5 , maka du = -6x dx atau
x dx = -1/6 du
dxxe x 53 2
9 +−
∫
CeCedue xuu
+−=+−=−= +−
∫
53 2
6
9
6
9
6
9
dxxe x 53 2
9 +−
∫
7. Hitunglah
Jawab
Misalkan u = cos x , maka du = -sin x dx atau
sin x dx = -du.
Sehingga
xdxtan∫
dx
x
x
xdx ∫∫ =
cos
sin
tan
CxCxCu
u
du
dx
x
x
xdx +=+−=+−=
−
== ∫∫∫ seclncoslnln
cos
sin
tan
9. Bentuk integral dapat
diselesaikan dengan metode Integral By Parts
(Integral sebagian – sebagian) , yaitu
dxxfxgxgxfdxxgxf ∫∫ −= )(')()()()(')(
dxxgxf∫ )()(
Atau lebih dikenal dengan rumus
duvuvdvu ∫∫ −=
10. Hitunglah
Jawab
Misalkan u = 3 – 5x , du = -5 dx.
dv = cos 4x , v = ¼ sin 4x dx
Maka
dxxx )4cos()53(∫ −
∫∫ −−−=− )5)(4sin()4sin()(53()4cos()53( 4
1
4
1
dxxxxdxxx
11. Hitunglah dxxx )ln()5( 3
∫ +
dxxe x
)cos(2
∫
dxxx )4cos(2
∫
a
b
c
Exercise
13. The method of Partial Fractions provides a way
to integrate all rational functions. Recall that a
rational function is a function of the form
where P and Q are polynomials.
1. The technique requires that the degree of the
numerator (pembilang) be less than the degree
of the denominator (penyebut)
If this is not the case then we first must divide
the numerator into the denominator.
dx
xQ
xP
∫ )(
)(
14. 2. We factor the denominator Q into powers of
distinct linear terms and powers of distinct
quadratic polynomials which do not have real
roots.
3. If r is a real root of order k of Q, then the partial
fraction expansion of P/Q contains a term of the
form
where A1, A2, ..., Ak are unknown constants.
k
k
rx
A
rx
A
rx
A
)()()( 2
21
−
++
−
+
−
15. 4. If Q has a quadratic factor ax2
+ bx + c which
corresponds to a complex root of order k, then the
partial fraction expansion of P/Q contains a term of
the form
where B1, B2, ..., Bk and C1, C2, ..., Ck are
unknown constants.
5. After determining the partial fraction expansion of
P/Q, we set P/Q equal to the sum of the terms of
the partial fraction expansion. (See Ex-2.Int.Frac)
k
kk
cbxax
CxB
cbxax
CxB
cbxax
CxB
)()( 222
22
2
11
++
+
++
++
+
+
++
+
16. 6. We then multiply both sides by Q to get some
expression which is equal to P.
7. Now, we use the property that two polynomials
are equal if and only if the corresponding
coefficients are equal.
(see ex3-int.Fractional)
8. We express the integral of P/Q as the sum of
the integrals of the terms of the partial fraction
expansion.
(see Ex4-Int.Fractional)
17. 9. Integrate linear factors:
rxAdx
rx
A
−=
−∫ ln
)(
1
1
111
)(
1)(
+−
−
+−
=
−∫
n
n
rx
n
A
dx
rx
A
for n > 1
18. 10. Integrate quadratic factors:
Some simple formulas:
++=
+
+
∫ a
x
A
C
ax
B
dx
ax
CBx
arctan)ln(
2
22
22
+
+
−
=
+
+
∫ a
x
a
C
axa
BaCx
dx
ax
CBx
arctan
2)(2)( 3222
2
222