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Description

permacht asrTM
integral Calculus
theorems types of iNtegrals
averaGe value Theorem indeFiniTe inTeGralS
• If f(x) is continuous over [a,b], then • If F(x) + C has the derivative f(x), then F(x) + C is the anti-derivative or the indefinite integral of f(x)
1 b
f(x) = f(x) dx = y Given d d[ ]x) + Cf= =() F erivative,then [ ]) + CC f= ()d xnti-derivative or indefinite integral
ba − ∫a dx ∫
which means that the integral
where: c = integration constant f(x) = integrand F(x) = particular integral
equals the area of the rectangle
(b)− =•(fx )( ba − )• y d 4 3 4 3
example: Given d C+ =4=x x erivative,then + C= ∫xd x = anti-derivative or indefinite integral
f(x) dx
y= f(x)
deFiniTe inTeGralS
definition 1 Givent he indefiniteintegral f(x) dx = F(x)+ C
∫
then bf()dx = +Fb () CF − +()aC = −() F (a)
a x b ∫a [ ] [ ]
b
wheref ∫a ()d x = d aefiniteintegral from to bfof (x)d x
FundamenTal TheoremS
(a,b = lower and upper limits of integration x = variable of integration)
• If f is continuous on [a, b], then
x definition 2 The limit of the sum of inscribed or circumscribed rectangles
F(x) = ∫a f(t) dt equals the area under f(x) between a and b and is given by the f(x) f(x)
d x definite integral between a and b
means ∫a () dt = ()x n b
dx lim ∑ f(xi x i= ∫ ()d x
n→∞ 1 a
(that is, F is the anti-derivative of f) x x x x
• If F is the anti-derivative of f, then a
b
b b
∫a f(x) dx = []x) a = −(b) F(a)
douBle inTeGralS z y
• If f’ is continuous, then z = f(x,y)
x • The limit of the sum of inscribed or circumscribed prisms with base DA = DxDy
ft() dt = f(x) f(a) equals the volume under the surface z = f (x, y) and is given by the double y2
∫a
integral between (x , 1 ) 2nd (y , y1) 2
(that is, if you first differentiate f n x2 y2 y1
and then integrate from a to x, lim ∑ f(x,i i) ( i fx ,)yd ydx A
the result will differ from f(x) by at n→∞ 1 ∫x1 y1
x1
most f(a)) x2 x
• If a is chosen so that f(a) = 0, then Triple inTeGralS
differentiation and integration • Triple integrals are three-dimensional extensions of double integrals defined in analogous fashion
exactly cancel each other out n x y z
lim f(x, yz,) V =f 2 2 2 (, y ,)zdzdydx
averaGe value oF a n→∞ ∑1 i i i i ∫1 y1 ∫z1
FunCTion
iTeraTed & parTial inTeGralS
• If f(x) = continuous over [a, b], then x2 y2f(x, y) y(dx=f x2 y2 x,y) dy dx= 2 x2f(x, y) xd y
b • For evaluation purposes, double and ∫x1 y1 ∫x1∫ y1 ∫y1 x1
ydx
y = ∫a = average ordinate triple integrals are expressed in iterated double integral iterated integral iterated integral
x bdx with respect to forms of interchangeable order y2 x2
∫a ∫ ∫ f(x, ) y , f(x, y) xI = Partial Integrals
b abscissa x over [a,b] • Each iteration is called a partial integral y11 1
yds
y = ∫a = average ordinate
s b with respect to arc double integral Triple integral
∫ads
length s over [a,b] b 2 x() b h2z() 2 y(, z)
∫ ∫ f(x, ) ydx ∫ ∫ ∫ f(x, y,) dydxdz
a gx() a h1z()g1y(, z)
Evaluate in 2 Steps Evaluate in 3 Steps
y2 2 (x) x2 2 (y, z )
Note:Theprocesswhichis ∫ f(x, y) y ∫x g=(y, z )x, y,) dx
❶ y1 1(x) ❶ 1 1
theinverseofdifferentiationis
termed anti-differentiationor [at constant x → u(x)] [at constant y, z→ u(y, z)]
b y h= (x)
integration • Every continuous u(x) dx = U−(b) U(a) 2 2 u(y, z)d y
function f(x) has an ❷ ∫a ❷ ∫y1 1(x)
anti-derivativewhichisthe U = anti-derivative of u [at constant z → v(z)]
integral of f(x)
b
∫a v(z) dz = −(b) V(a)
❸
1 InTegral calculus • a-834-6 www.permacharts.com © 2009-2013 Mindsource Technologies Inc. permacht asr
areas By iNtegratioN Volumes & curVe leNgths
area under a Curve • Calculate volumes and curve lengths by single or multiple integrals
Y Y volumeS oF revoluTion
y2 y =f(x)x =g(y) r =hθ)
about x-axis about y-axis
Ay r
A θ method of x x y y
1 A θ V =yπ=π 2 2dx 2 2[(fx )] dx V =xπ=π 2 2dy 2 2 [( y )] dy
x θ2 1 cross sections x ∫x1 x1 y ∫y1 y1
x x
1 2 X X
x2 x2 1 θ2 2 2 1 θ2 method of y2 x2
A x= =∫x dx x f(x) dx A θ= = ∫ ∫ dθhθ [( )] dθ cylindrical V =y2π ∫y g(y) dy V =x2π ∫x f(x) dx
1 1 2 1 2 1 1 1
y2 y2 shells
A y= =∫y1 dy ∫y1 g(y) dy
SurFaCe oF revoluTion volumeS under a SurFaCe
arc revolving about x-axis rectangular Coordinates Cylindrical Coordinates
θ r g= (θ)
2 Y V =f x2 y2 2=(x)(, y )dydx V =g 2 2 2 (r,rθ)θdrd
x2 dy y= f(x) ∫x1 y1 1(x) θ1∫ 1 1=(θ)
A x= 2+π1∫x dx x= g(y)
1 dx
2 A volumeS aS Triple inTeGralS
y2 dx x
= +2π1∫yyy dy x1 x2 X
1 dy Z Z
θ ρsin
arc revolving about y-axis P θ P ρ
ρ
2 2 z θ
2 dx x2 dy θ
A y=2+π1∫y dy =2+1 x∫xx dx Y ρsin Y
1 dy 1 dx θ θ θ θ
Q ρρ Q(x,y)
area BeTween Two CurveS X X
x2 y2 rectangle cylindrical spherical
A =f ∫ ∫[ ]) g () d xA = − [ ]y) g(y) dy
x1 y1 V d xdydz V d ρρ dzdθ V =d ρ θsin ρdθd φ
∫∫∫ ∫∫∫ ∫∫∫
area oF a Curved SurFaCe
Z
rectangular Coordinates z= f(x,z= g(θ, ) lenGTh oF a Curve
www.permacharts.com
x2 y2 2 (x) ∂z 2 ∂z A rectangular Coordinates polar Coordinates
A = +∫ ∫∫ 1 + dydx
x1 y11= (x) ∂x ∂y Y 2 2 2 2
x= x S = +2 1 1dy dx = + 2 dx dy S =r + 2 dr d r= 2r 1+ 2 dθ dr
1 x1∫ ()dx y1 dy θ1 dθ ∫1 dr
y=2f (y= 1(x)
Cylindrical Coordinates x= x2
X example:
θ2 2 2 θ() ∂ z 2 ∂ z 2
A = + 1 + 1 rdrdθ • Find the volume of the solid
∫1∫ r11(θ) ∂r r2 ∂θ generated by revolving
y = f(x) = 3x 2– x around the
example y-axis, with x 1 0, x =23 3
2 3 23 3 4 1 5 243
• Find the area in the first octant cut from the cylindrical V=x2π∫x∫ f(x) dx 0x( )2− x=d xV π x − x = π
surface z = 4 – x 2 by the plane y = x 1 4 5 0 10
2 3 3 1 3 243
• Since boundaries of projection of the required∫ ∫ea onto the x( )2− x=d xV π x − x 5 = π
x1 0 4 5 0 10
x – y plane are y = 0, y = x, and x = 2, we have
z 2
2 x ∂z 2 ∂z2 z = 4–x
A = d∫0 0 1 (∂x + ()∂y ydx y = x
2 x
= + 14 x 2 + 0dydx y
∫0 0
2 1 iNdefiNite iNtegrals
A =x+ 14 2xdx = − 1( )7 1
∫0 12
• Frequently occurring indefinite integrals
2
• This result can also be found, though x [integration by parts]
with much greater difficulty, by ∫udvu∫=−vv ∫ du+C u or ∫ vd′xu =−v v u′ x+C
reversing the order of integration: ∫[a]u+∫ =v dx au∫dx +b+v d∫ C x ln dx = x−lxn xC+
2 2
A =x+ 14 2 + 0dxdy 1
∫0 y ∫ ∫dx= +lnC sin xdx= − cosC ∫ cos sdx= x in C
x
• Thus, ease of evaluation f′()) bax ax
dx =2+ f(x) C b axdx = + C e axdx= e +C
often depends on order ∫ ∫f(x) aln ∫ a
of integration
f′() g(y)
∫ dx=ln+[(fx )] C g ∫ ∫()d x= ∫ y′ dy + C a dx=a+x C
f(x)
2 2 n 1
xx dx= − x lnx x + C x ndx x + ≠n –1
∫ 2 4 ∫ n +1
2 InTegral calculus •a-834-6 www.permacharts.com © 2009-2013 Mindsource Technologies Inc. permacht asr
momeNts, ceNtroids & momeNts of iNertia
Moment M = mass of a particle x distance from an axis or plane example y y = 4 – x
Moment of Inertia I = mass of a particle x distance squared from an axis or plane 4
• A solid of revolution is generated
Centroid x, y, z = center of gravity r = density = 1 by revolving the parabola
y = r – x2in the x – z plane
moment Centroid moment of inertia (i.e., about the y-axis)
• Compute moment, centroid, x
plane revolution –2 +2
M yx dA 1 Ixy= 2dA and moment of in

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