University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 380 Group G1 Graded Homework V Correction. 1. Compute the derivative of the function x 7? tan?1(x) = arctan(x) ; use it to compute ∫ b a dx x2 + 1 , where a, b ? R (in terms of arctan(a), arctan(b)), then to compute ∫ 1 0 dx x2 + x + 1 . With a change of variable, compute the integral ∫ pi 2 0 cos(x)dx 2? cos2(x) + sin(x) . Correction. A direct computation shows that tan?(x) = 1 + tan2(x). The fact that tan(arctan(x)) = x, and the Chain Rule for functions of one real variable, yields tan?(arctan(x)). arctan?(x) = 1, so that arctan?(x) = 1 tan?(arctan(x)) = 1 1 + (tan(arctan(x)))2 = 1 1 + x2 . This immediately yields ∫ b a dx x2 + 1 = arctan(b)? arctan(a). One has ∫ 1 0 dx x2 + x + 1 = ∫ 1 0 dx (x + 12 ) 2 + 34 = ∫ 3 2 1 2 dy y2 + 34

• urbana-champaign fall


• theorem gives


• dx x2


• real variable


• variable

Sujets

• Function of a real variable
• Variable

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 11
• Langue: English

Extrait

Z b dx−11. x 7→ tan (x) = arctan(x)
2x +1aZ 1
dx
a,b∈R arctan(a),arctan(b) 2x +x+10Z π
2 cos(x)dx
22−cos (x)+sin(x)0
0 2tan (x) = 1 + tan (x) tan(arctan(x)) = x
0 0tan (arctan(x)).arctan (x) = 1
1 1 10
arctan (x) = = = .0 2 2tan (arctan(x)) 1+(tan(arctan(x))) 1+x
Z b dx
= arctan(b)−arctan(a)
2x +1a
√ 3Z Z Z 3 Z 31 1 22 2dx dx dy 4 dy 4 3 2
√= = = = arctan( y)2 1 3 3 22 2 2x +x+1 1 3 1 √ 3 2(x+ ) + y + 1+( y) 3 10 0 2 4 42 2 3 2
√ π 1 π
arctan( 3) = arctan(√ ) =
3 63
Z 1 dx 2 π π
= √ = √ .
2x +x+1 63 3 30
π
u = sin(x) du = cos(x)dx x7→ u(x) [0, ] [0,1]
2
Z π Z Z1 12 cos(x)dx du du π
√= = = .
2 2 22−cos (x)+sin(x) 2−(1−u )+u u +u+1 3 30 0 0
2. D
2 3D x≥ 0 y≥ 0 y = x y = x
2/3 2/3D x,y≥ 0 x +y ≤ 1
1 1/3
3u = x v = y
πZ
2 π2 2sin (θ)cos (θ)dθ =
160
0≤ x≤ 1 0≤ y≤ 1
3/2
x ≤ x D
ZZ Z Z Z 11 x 1 2x 2 1 2 13/2 5/2dxdy = dy dx = (x−x )dx = − x = − = .
3/2 2 5 2 5 10D x=0 y=x x=0 0
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∂(u,v)2 2(x,y) 7→ (u,v) R R
∂(x,y)
1 1−2/3 −2/3x y = (u,v) R
2 29 9u v
+C
ZZ ZZ ZZ
∂(x,y) 2 2(D) = dxdy = | |dudv = 9u v dudv .
+ ∂(u,v) +D C C
u = ρcos(θ)
v = ρsin(θ)
π πZ Z Z1
2 294 2 2 2 2(F) = 9ρ sin (θ)cos (θ)ρdρ dθ = sin (θ)cos (θ)dθ .
6θ=0 ρ=0 0
2sin (2θ) −cos(4θ)+12 2sin (θ)cos (θ) = =
4 8
πZ
2 π 3π2 2sin (θ)cos (θ)dθ = (D) =
16 320
ZZ
3. f(x,y)dxdy
D
x+y 2f(x,y) = e D ={(x,y)∈R :|x−y|≤ 1,|x+y| < 1}
2 2x y2f(x,y) = x −2y D + = 1
2 2a b
2 2f(x,y) = x +y −2y D (1,1) 1
2y2
f(x,y) = xy D (x,y) x,y≥ 0 x + ≤ 1
4
u = x+y v = x−y
(u,v) −1 < u < 1 1 ≤ v ≤ 1 (x,y) 7→ (u(x,y),v(x,y))
u+v u−v ∂(x,y)
x = y = | |
2 2 ∂(u,v)
1 −1 1 1 1
x y (u,v) | . − | =
2 2 2 2 2
ZZ Z Z Z1 1 11 1u uf(x,y)dxdy = e dudv = e du = e− .
2 eD u=−1 v=−1 u=−1
x = arcos(θ) y = brsin(θ)
r 0 1 θ 0
2π

∂(x,y) acos(θ) −arsin(θ) 2 2= = abrcos (θ)+abrsin (θ) = abr .
bsin(θ) brcos(θ)∂(r,θ)
a b
ZZ Z Z Z2π 1 2π 3 2 a b 2ab2 2 2 2f(x,y)dxdy = (abr)(a r cos (θ)−2brsin(θ))dr dθ = cos (θ)− sin(θ) dθ .
4 3D θ=0 r=0 θ=0
0
2cos(θ)+12cos (θ) =
2
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4
2 2cos (θ)+sin (θ) = 1
x− 1 = rcos(θ) y− 1 = rsin(θ) 0 ≤ r ≤ 1
0≤ θ≤ 2π
ZZ Z Z Z 2π 1 2π 1 2 π2f(x,y)dxdy = (r +2rcos(θ))rdθ dr = + cos(θ) dθ = .
4 3 2D θ=0 r=0 θ=0
√√ 2ZZ Z Z 2 Z Z4−4x1 4−4x 1 12 2y x(4−4x ) 1 1
f(x,y)dxdy = xydy dx = x dx = dx = 1− = .
2 2 2 2D x=0 y=0 x=0 00
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