Penerapan Fungsi Eksponen Dan LogaritmaFull description
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rumus grammarDeskripsi lengkap
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TRIGONOMETRIFull description
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Rumus-rumus pada file ini: - Faktor-faktor Konstanta dan Konversi - Satuan-satuan dan Simbol-simbolnya - Prefiks - Nilai dari Fungsi Trigonometri untuk Sudut-sudut yang Umum - Mekanika Newton ...
Rumus dasar integral fungsi eksponen au
1.
∫
2.
∫ e
au
=
u
ln a
+c , a >0, a≠13
= eu +c
Contoh : Selesaikanlah integral fungsi eksponen berikut ini : 1. ∫ e−x dx u = −x
mis
= −dx
du maka: −x
∫ e 2.
x
= − ∫ eu dx = −eu +c = −e−x +c
dx
+1)3 ex dx
∫ (e
u = (ex +1)
mis
du = ex dx maka: 3 x ( ) + e 1 ex dx ∫
3.
∫ x
2
∫ u
3
=
dx =
1 4 u +c 4
=
1 (ex +1)4 4
+c
−x3 dx
e
u = − x3
mis
du = − 3x2 dx maka: 2
∫ x 4.
3
3
e−x dx = − 1 eu du = − 1 eu + c = − 1 e−x + c
2 − 3x
∫ x e
3 ∫
3
3
dx
u = − 3x2
mis
du = − 6xdx maka: 2
2
x e−3x dx = − 1 eu du = − 1 eu + c = − 1 e−3x + c
∫ 5.
6
∫ (x +3) mis
2
ex
∫
6
6
+6x dx
u = x2 + 6x du = (2x + 6) dx = 2(x + 3) dx
maka: x2 + 6x
∫ (x+ 3) e
2 dx = 1 eu du = 1 eu + c = 1 ex + 6x + c 2
∫
2
2
1
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Contoh intergal parsial Selesaikanlah integral parsial berikut ini : 1.
∫ x ex dx = x du = dx
mis
u
dv v
= ex dx = ex
maka : x
∫ xe 2.
∫ x3
dx
= xex −∫ ex dx
= x ex − ex +c
e2x dx
u = x3
mis
du = 3x2 dx dv = e2x dx v = 1 e2x 2
maka: 3
∫ x
mis
e2x dx = x3 1 e2x − 1 e2x 3x2 dx = 2
u = x2 du = 2xdx dv = e2x dx v = 1 e2x 2
∫ 2
1 3 2x x e −3 2 2
2
∫ x
e2x dx
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maka : 3
∫ x
2x
e
dx
1 3 2x x e 2
=
[
1 3 2x x e 2
− ∫ xe2x dx
=
1 3 2x 3 2 2x x e − x e 2 4
+
2x 3 e x dx 2
− 3 x2 e2x +
2x 3 e x dx 2
du
dv
=
e
v
=
1 2x e 2
−3 2
[x
2 1 2x e 2
2
∫
dx
2x
dx
maka : 3
∫ x
3.
x
∫ e
2x
e
dx
=
1 3 2x x e 2
=
3 2 2x 1 3 2x x e − x e 2 4
+
3 1 2x x e 2 2
=
3 2 2x 1 3 2x x e − x e 2 4
+
2x 3 1 xe 2 2
=
3 2 2x 1 3 2x x e − x e 2 4
+
2x 3 xe 4
− 3 ∫ e2x dx
=
3 2 2x 1 3 2x x e − x e 2 4
+
2x 3 xe 4
− 3 e2x + c
4
∫
[ [
]
− ∫ 1 e2x dx 2
]
− 1 ∫ e2x dx 2
4
8
sin x dx
mis
u
= ex
du
= exdx
dv
= sin x dx
v
= −cos x
maka : x
∫ e
mis
sin x
u
dx
= ex cos x −∫ −cos x ex dx
= ex
du
= exdx
dv
= cos x dx
v
= sin x
]
− ∫ 1 e2x 2x dx
]
3 1 2 2x 1 3 2x x e − x e 2 2 2
x
u
2
=
= =
mis
− 3 ∫ x2 e2x dx =
= ex cos x +∫ cos x ex dx
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maka: x
∫ e
[
]
sin x dx = ex cos x + ex sin x − ∫ sinx ex dx
= ex cos x + ex sin x − ∫ sin x ex dx = ex cos x + ex sin x − ∫ ex sinx dx selanjutnya x
∫ e
sin x dx +
x
∫ e
sin x dx = = ex cos x + ex sin x
selanjutnya 2∫ ex sin x dx = ex cos x + ex sinx sehingga x
∫ e
(
)
sinx dx = 1 ex cos x + ex sinx 2
=
1 x e ( 2
cos x + sinx)
TUGAS 6 Selesaikanlah integral berikut ini dengan metode parsial : 1. ∫ et sin t dt 2.