Examen 3 (solutions) 17 avril 2007
Q.1. Points d’intersection : x3= 4x2−3x⇒x3−4x2+ 3x=x(x−1)(x−3) = donc x = {0,1,3}. L’aire est :
Z 0
−1
−x3+4x2−3x dx+
Z 1
0
x3−4x2+3x dx+
Z 3
1
−x3+4x2−3x dx+
Z 4
3
x3−4x2+3x dx
=−x4 4 +4x3
3 −3x2 2
0
−1
+x4 4 −4x3
3 +3x2 2
1
0
+−x4 4 +4x3
3 −3x2 2
3
1
+x4 4 −4x3
3 +3x2 2
4
3
=44−2(3)3+ 3
4 +8(3)3−4(4)3−4
3 +3(4)2−6(3)2+ 9
2 =205
4 −44 3 +3
2 =457 12 Q.2. a) π
Z 2
−1
(x2+ 1)2 dx= π Z 2
−1
x4+ 2x2+ 1dx=π x5
5 +2x3 3 +x
2
−1
π 25
5 +24 3 + 2
−π
−1 5−2
3 −1
= 78π 5 b) π
Z π4
0
2
√2 2
dy−π Z π4
0
sec2y dy= 2πy
π 4
0
−πtany
π 4
0
=π2 2 −π
c) 2π Z 12
0
(x+ 2)(3−x2−ex)dx= 2π Z 12
0
−xex−ex−x3−2x2+ 3x+ 6dx
= 2π
−xex+ex−ex−x4 4 −2x3
3 +3x2 2 + 6x
1 2
0
= 2π 629 192−e12
2
!
Q.3.
Z 1
−3
s 1 +
ex−e−x 2
2
dx= Z 1
−3
r 1 + e2x
4 −1
2 +e−2x 4 dx=
Z 1
−3
re2x 4 +1
2 +e−2x 4 dx Z 1
−3
s
ex+e−x 2
2
dx= 1
2(ex−e−x)
1
−3
=1 2
e−1
e − 1 e3 +e3
Q.4. a) Z 1
−2
4x+ 6
(x2+ 1)(x+ 2) dx= lim
a→−2+
Z 1
a 2 5x+3210
x2+ 1 + −2 5(x+ 2)dx lim
a→−2+
1
5ln|x2+ 1|+32
10arctanx−2
5ln|x+ 2|
1
a
=−∞(impropre, diverge)
b) Z π2
0
tanx dx= lim
a→π2−
Z a
0
tanx dx= lim
a→π2−(−ln|cosx|)
a
0
=∞(impropre, diverge)
c) Z ∞
−∞
dx
1
2+x2 = lim
N→−∞
1 2
Z 0
N
dx 1 + (√
2x)2 + lim
M→∞
1 2
Z M
0
dx 1 + (√
2x)2
= lim
N→−∞
1
2arctan(√ 2x)
0
N
+ lim
M→∞
1
2arctan(√ 2x)
M
0
= π
2 (impropre, converge)
