Sabtu, Juli 28, 2012

TEKNIK PENGINTEGRALAN 02

Melanjutkan beberapa contoh-contoh soal dari Integral dan solusinya :
CONTOH :
$\dpi{100} \fn_cs \int x^{2}.\sqrt{x-2}.dx=....$
Integral diatas bisa saja diselesaikan dengan INTEGRAL PARTIAL
tp menurut aq penyelesaian pake partiaL adalah senjata pamungkas, kalo bisa pake
substitusi ....why not ???
Oke-oke kita jelaskan dulu dech gmana sich Integral Partial itu.
ada buku yang menyatakan kita pake partial jika pada soal terdapat
dua fungsi berbeda yang dikalikan seperti
$\dpi{100} \fn_cs {\color{Red} x^{2}}.{\color{Blue} \sqrt{x-2}}$ ..................yang merah fungsi kuadrat dan yang biru fungsi irasional
$\dpi{100} \fn_cs {\color{Red} x^{2}}.{\color{Blue} sin\, (4x)}$ ................yang merah fungsi kuadrat dan yang biru fungsi trigonometri
$\dpi{100} \fn_cs {\color{Red} e^{3x}}.{\color{Blue} sin\, (4x-\pi )}$ ........yang merah fungsi eksponendan yang biru fungsi trigonometri......etc

Sementara cara pengerjaannya bisa pake rumus dan bisa juga
memakai TABEL TANZALIN....Nach dah keliatan kan ......<g tau apanya ????>

Jumat, Juli 27, 2012

TEKNIK PENGINTEGRALAN

Si Mawar (=bukan nama asline) pelajar kelas XII - IPA suatu SMA.
dia bingung banget karena di sekolah gurunya nerangin bab integral super kilat banget,
belum selesai dia mengerti cara penyelesaian yg satu timbul cara laen.
Ea....mungkin Mawar adalah salah satu dari ribuan ribuan siswa SMA yang Galau akan
bab integral....si Mawar tau bener bahwa Bab di kelas XII lumayan banyak dan dia juga
tau klo kelas XII di UNAS keluar 40% dari jumlah soal dan dia juga telah ngintip soal2 SNMPTN tulis
taon 2012 ini juga banyak yg kelas XII.

Nah sekarang aq berikan aja alur perpikir klo kita punya masalah INTEGRAL

Kalo pgen punya file alur diatas bisa download disini

Kita ambil beberapa contoh :

CONTOH 1 :
$\dpi{100} \fn_jvn \int {\color{Red} 8x}.({\color{Blue} 4x^{2}+6})^{10}.dx$
kita lihat yang berwarna merah dan biru
yang biru adalah bapak n yang merah adalah mengandung turunannya (anak)
jadi dalam soal terdapat FUNGSI beserta turunannya
ATAU ADA YANG BILANG SELISIH PANGKATNYA ADALAH SATU
$\dpi{100} \fn_jvn \int {\color{Red} 8x}.({\color{Blue} 4x^{2}+6})^{10}.\frac{d({\color{Blue} 4x^{2}+6})}{8x}=\int ({\color{Blue} 4x^{2}+6})^{10}.d({\color{Blue} 4x^{2}+6})$
$\dpi{100} \fn_jvn =\frac{1}{11}.(4x^{2}+6)^{11}+C$

CONTOH 2 :
$\dpi{100} \fn_jvn \int {\color{Red} 6x}.({\color{Blue} 3x+6})^{7}.dx$
khusus untuk pangkat satu boleh g berselihnya satu alias sama....
qt masih bisa pake integral substitusi kok.....caranya adalah :
$\dpi{100} \fn_jvn Misal\:\: p=3x+6\Rightarrow 3x=p-6\: \: dan\: \: 6x =2p-12$
$\dpi{100} \fn_jvn p=3x+6\Rightarrow \frac{dp}{dx}=3\: \: atau\, dapat \, ditulis\: \: \frac{dp}{3}=dx$
jadi soal kamu sekarang menjadi :
$\dpi{100} \fn_jvn =\int (2p-12).p^{7}.\frac{dp}{3}=\frac{1}{3}.\int (2p^{8}-12p^{7}).dp$
Nach kalo sudah jadi gni ......bsa kan ngelanjutin'e.....
n jangan lupa trus p nya dikembaliin lagi jadi (3x +6)

Senin, Juli 23, 2012

BELAJAR INTEGRAL TRIGONOMETRI LAGI

Sekarang saatnya kita belajar lewat Video yuk
SOALNYA :
$\dpi{100} \fn_phv \int x.cos\, (x^{2}+1).dx=....$
dan dibawah ini penyelesaiannya .

SOALNYA :
$\dpi{100} \fn_phv \int sin^{3}x.cos\, x.dx=....$
dan dibawah ini penyelesaiannya

SOALNYA :
$\fn_phv \int_{0}^{\frac{\pi }{2}} x.cos\, x.dx=....$
dan dibawah ini penyelesaiannya

SOALNYA :
$\fn_phv \int \left (a.cos\, x + \frac{b}{sin\, ^{2}x}\right ).dx=....$
dan dibawah ini penyelesaiannya

yang lain-lain dapat kamu kunjungi alamat disini prens

Minggu, Juli 22, 2012

BERMAIN-MAIN DENGAN INTEGRAL 3GONOME3

List of integrals of trigonometric functions

Diambil langsung dari http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions

untuk belajar integral trigonometri nt bisa download di sini
Nich salah satu tampilane

SOAL n PEMBAHASAN INTEGRAL_5

NO-1
$\dpi{80} \fn_jvn Jika \:\: \: a =sin^{2}15^{0}\: \: dan\: \: b =cos^{2}15^{0}\: \: maka\: \:\int_{a}^{b}(x-a)(x-b).dx=$
$\dpi{80} \fn_jvn A.-\frac{\sqrt{3}}{16}$ (*)
$\dpi{80} \fn_jvn B.-\frac{\sqrt{3}}{8}$
$\dpi{80} \fn_jvn C.-\frac{\sqrt{3}}{6}$
$\dpi{80} \fn_jvn D.-\frac{\sqrt{3}}{4}$
$\dpi{80} \fn_jvn E.-\frac{\sqrt{3}}{2}$

dengan bantuan WolframAlfa
akan lebih asyik....
karena kita bisa langsung
masukin hasilnya
Bentuk soal diatas akan lebih mudah kita kerjakan dengan memakai INTEGRAL PARTIAL.
yaitu memakai rumus atau pakai tabel TANZALIN.
$\dpi{100} \fn_jvn Rumus\: {\color{Magenta} integral\: partial}\: \int u.dv=u.v-\int v.du$
$\dpi{100} \fn_jvn u=(x-a)\Rightarrow du=dx$
$\dpi{100} \fn_jvn dv=(x-b).dx\Rightarrow v=\frac{1}{2}(x-b)^{2}$
$\dpi{100} \fn_jvn (x-a).(x-b)^{2}-\int \frac{1}{2}(x-b)^{2}.dx$
$\dpi{100} \fn_jvn (x-a).(x-b)^{2}- \frac{1}{2}.\frac{1}{3}(x-b)^{3}\mid _{a}^{b}$ $\dpi{100} \fn_jvn =\frac{1}{6}(a-b)^{3}$
$\dpi{100} \fn_jvn =\frac{1}{6}(sin^{2}15^{0}-cos^{2}15^{0})^{3}=-\frac{1}{6}.(cos^{2}15^{0}-sin^{2}15^{0})^{3}$
$\dpi{100} \fn_jvn -\frac{1}{6}.(cos\, 30^{0})^{3}=-\frac{1}{6}.(\frac{\sqrt{3}}{2})^{3}=-\frac{1}{6}.(\frac{3\sqrt{3}}{8})=-\frac{\sqrt{3}}{16}$

Beberapa contoh soal yang memakai teknik INTEGRAL PARTIAL
bisa di simak dibawah ini yang aq ambilkan di alamat http://www.intmath.com/blog/tanzalin-method-for-easier-integration-by-parts/4339

Example 1

$\int2x(3x-2)^6dx$

Integration by Parts Method

First, let’s see normal Integration by Parts for comparison.

We identify u, v, du and dv as follows:

u = 2x	dv = (3x − 2)⁶dx
du = 2dx	$v=\frac{1}{21}(3x-2)^7$

Integration by Parts then gives us:

$\int{udv}=uv-\int{vdu}$

$\int2x(3x-2)^6dx=\frac{2x}{21}(3x-2)^7-\frac{2}{21}\int(3x-2)^7dx$

Now, we find the unknown integral:

$\int(3x-2)^7dx=\frac{1}{24}(3x-2)^8+C$

Putting it together, we have:

$\int2x(3x-2)^6dx=\frac{2x}{21}(3x-2)^7-\frac{1}{252}(3x-2)^8+C$

We can then factor and simplify this to give:

$\int2x(3x-2)^6dx=\frac{21x+2}{252}(3x-2)^7+C$

The Tanzalin Method is somewhat less messy.

Example 1, now using Tanzalin Method

In the Tanzalin Method, we set up a table as follows. In the first column are successive derivatives of the simplest polynomial term of our integral. (We need to choose this term for the derivatives column because it will disappear after a few steps.)

In the second column are the integrals of the second term of the integral.

We just multiply the 2 terms with green background in the table (the original 2x term and the first integral term). We don’t change the sign of this term.

We then multiply the 2 terms with yellow background (the first derivative and the second integral term). We assign a negative sign to the product, as shown.

The answer for the integral is just the sum of the 2 terms in the final column.

The question again, for reference:

$\int2x(3x-2)^6dx$

Derivatives	Integrals	Sign	Same-color Products
2x	(3x − 2)⁶
2	$\frac{1}{21}(3x-2)^7$	+	$\frac{2x}{21}(3x-2)^7$
0	$\frac{1}{504}(3x-2)^8$	−	$-\frac{1}{252}(3x-2)^8$

Summing the 4th column:

$\int2x(3x-2)^6dx=\frac{2x}{21}(3x-2)^7-\frac{1}{252}(3x-2)^8+C$

(We add the constant of integration, C only at the end, not in the table.)

We can then factor and simplify this to give:

$\int2x(3x-2)^6dx=\frac{21x+2}{252}(3x-2)^7+C$

Example 2

$\int{x}\sin{x}\hspace{3}dx$

We’ll go straight to the Tanzalin Method.

Derivatives	Integrals	Sign	Same-color Products
x	sin x
1	−cos x	+	−x cos x
0	−sin x	−	sin x

We multiplied (x) by (−cos x) and we didn’t change the sign.

We then multiplied (1) by (−sin x) and changed the sign.

Adding the final column gives us the answer:

$\int{x}\sin{x}\hspace{3}dx=-x\cos{x}+\sin{x}+C$

Example 3

$\int{x^2}\sqrt{x-1}\hspace{3}dx$

Using the Tanzalin Method requires 4 rows in the table this time, since there is one more derivative to find in this case.

We need to alternate the signs (3rd column), so our 4th row will have a positive sign.

Derivatives	Integrals	Sign	Same-color Products
x²	$(x-1)^\frac{1}{2}$
2x	$\frac{2}{3}(x-1)^\frac{3}{2}$	+	$\frac{2x^2}{3}(x-1)^\frac{3}{2}$
2	$\large{\frac{4}{15}(x-1)^{\frac{5}{2}}}$	−	$\large{-\frac{8x}{15}(x-1)^{\frac{5}{2}}}$
0	$\large{\frac{8}{105}(x-1)^{\frac{7}{2}}}$	+	$\large{\frac{16}{105}(x-1)^{\frac{7}{2}}}$

So our final answer is:

${\int{x^2\sqrt{x-1}}dx=}{\frac{2x^2}{3}(x-1)^{\frac{3}{2}}-}{\frac{8x}{15}(x-1)^{\frac{5}{2}}+}{\frac{16}{105}(x-1)^{\frac{7}{2}}}+C$

Example 4 – a Problem Arises

This is the same question as Example 3 in the Integration by Partssection in IntMath.

$\int{x^2}\ln{4x}\hspace{4}dx$

We need to choose ln 4x (natural logarithm of 4x) for the first column this time, following the Integration by Parts priority recommendations of:

log of x,
e raised to power x
x raised to a power

[Note: If we choose the other way round, we would have to find integrals of ln 4x, which is not pretty (and certainly no easier than doing it all using Integration by Parts. See Example 6 on this page:Integration by Parts).

Derivatives	Integrals	Sign	Same-color Products
ln 4x	x²
$\frac{1}{x}$	$\frac{x^3}{3}$	+	$\frac{x^3\ln4x}{3}$
$-\frac{1}{x^2}$	$\frac{x^4}{12}$	−	$-\frac{x^3}{12}$
$\frac{2}{x^3}$	$\frac{x^5}{60}$	+	$-\frac{x^3}{60}$

When do we stop? The derivatives column will continue to grow, as will the integrals column. The Tanzalin Method requires one of the columns to "disappear" (have value 0) so we have somewhere to stop.

So our final answer is:

$\int{x^2}\ln{4x}\hspace{4}dx=\frac{x^3\ln4x}{3}-\left(\frac{1}{12}+\frac{1}{60}+\frac{1}{180}+\frac{1}{420}+\ldots\right)x^3+C$

That expression in brackets must equal $\frac{1}{9}$ (since this is the answer we got using Integration by Parts), but as you can see, it is not a Geometric Progression and would take some figuring out.

Beberapa contoh integral 3Gonome3 dan langkah pemecahan dapat langsung
kunjungi alamat disini bro.
Dan tabel untuk integral-integral 3gonometri dapat di lihat di alamat sini Bro

Aku n MATEMATIKA cah SMA Nda

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