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Reference Guide

# Integral Calculus - Reference Guides

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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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