Intégration sur un segment (1/2)

1
0(1 −t2)ndt >1 0(1 −t)ndt 1 0(1 −t2)ndt
n16√1 + xn61 + (1 + 1
n)n[1; 1 + 1 n]
un=1+1/n
1√1 + xndx
f[a;b]g[a;b]∃c∈
[a;b],b
af(x)g(x)dx =f(c)b ag(x)dx
f[a;b]b
af(x)dx =b a|f(x)|dx ⇒[(f f )]
f[0; 1] 1
0f(x)dx =1
2c f(c) = c
a)un=1
nsin π n+ sin 2π n+···+ sin nπ nb)un=1
√n√1 + √2 + ···+√n c)un=
k=n
k=1
1 + k
n
1
n
d)un=n1
(n+ 1)2+1 (n+ 2)2+···+1 (n+n)2e)un=1 nr+1 (1r+ 2r+···+nr)r∈Qr>0
a)1
0(ex+x2
2−ln(1 + x))dx e +1 6−ln(4) b)1
0x(x+ 2 −e)exdx 0
c)1
0
x−2
(2x−3)2dx (−1 4ln(3) −1 6)d)π/4
0cos4xsin2xdx (1
16 (π 4+1 3))
e)2
1
x3
(1 + x4)2dx (15 136 )
n In=1
0xn√1−xdx I0I1∀n∈N∗,(3 + 2n)In= 2nIn−1
f f :x→−1x < 0
1x > 0
0
f1(x) = 1+x
1−xf(t)dt f1f1
f2(x) = 1+x
1−xf1(t)dt f2R
n In=π/2
0(sin x)ndx
I0I1InIn−2In
n
(In) lim
n→+∞
In−1
In
∀n>1, nInIn−1=π
2lim
n→+∞
Inlim
n→+∞
In√n
lim
n→+∞2n(I2n)2=π
2
lim
n→+∞
n1.3.5. . . . (2n−1)
2.4.6...2n
2
=1
π
u v Cn
b
a
u(x)v(n)(x)dx = (uv(n−1) −u0v(n−2) +···+ (−1)n−1u(n−1)v)(x)b
a+ (−1)nb
a
u(n)(x)v(x)dx
x2e3xdx π
0x3cos xdx
Pnn QnPn(x)eωxdx =
Qn(x)eωx
a)x
√1−x2exdx x−√1−x2
2ex+K)b)1
0ln(1 + x2)dx π
2+ ln(2) −2)
c)2
1
x
√1 + xdx 2√2
3)d)xdx
√x+ 1√x+ 3
e)1
0
x
1 + √1 + xdx 4√2
3−5 3)f)5
1
√x−1
xdx ( 4 −2 (2))
g)1
0
dx
x√1 + x2h)x
1
1−√t
tdt (−4 3(1 −√x)3/2)