Math bac ex 8

Page 1

‫اﻟﺘﻜﺎﻣـــــﻞ‬

‫اﻟﺜﺎﻧﻴﺔ ﺳﻠﻚ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ‬

‫ﺗﻤﺎرﻳﻦ و ﺣﻠﻮل‬ 4‫ﺗﻤﺮﻳﻦ‬

1 1 1 ‫ ﺗﺄآﺪ أن‬/‫ أ‬-1 − = t t + 2 t (t + 2)

dt

2

∫1 t ( t + 2 ) dt ‫ أﺣﺴﺐ‬/‫ب‬ π

∫02 e π

x

sin xdx

‫ أﺣﺴﺐ‬-2

π

J = ∫ x sin xdx ; I = ∫ 2 x 2 cos 2 xdx 2 0

2

2

‫ ﻧﻀﻊ‬-3

0

J ‫ و‬I ‫ ﺛﻢ اﺳﺘﻨﺘﺞ‬I − J ‫ و‬I + J ‫أﺣﺴﺐ‬

1 1 1 ‫ ﺗﺄآﺪ أن‬/‫ أ‬-1 − = t t + 2 t (t + 2) 1 1 t +2−t 1 − = = t t + 2 t (t + 2) t (t + 2) 2

∫1 2

∫1

dt dt t (t + 2)

‫ ﻧﺤﺴﺐ‬/‫ب‬

21 2 1 1 3 dt = ∫ − dt = ln t − ln ( t + 2 ) 1 = ln 2 − ln 4 + ln 3 = ln 1 t t (t + 2) t+2 2

π

A = ∫ 2 e x sin xdx 0

π

‫ ﻧﺤﺴﺐ‬-2

π

π

π

A = e x sin x  2 − ∫ 2 e x cos xdx = e x sin x  2 −  e x cos x  2 − K 0 0 0 0 π

π

1 A =  e x sin x  2 −  e x cos x  2 0 0 2  π

  = .........  π

J = ∫ 2 x sin xdx ; I = ∫ 2 x 2 cos 2 xdx 2

2

0

J +I =∫

π

2 x2 sin 2 0

xdx + ∫

π

2 x2 cos2 0

xdx = ∫

0

π

2 x2 0

( cos

2

)

π

π

1 3  2 2 x2 dx =  3 x  0 0

I − J ‫ﻧﺤﺴﺐ‬

1

-3

I + J ‫ﻧﺤﺴﺐ‬

x + sin x dx = ∫ 2

‫ﻧﻀﻊ‬

=

π3 24


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