积分公式

Calculus Cheat Sheet

Integrals

Definitions

Definite Integral: Suppose f(x) is continuous Anti-Derivative : An anti-derivative of f(x) on [a,b]. Divide [a,b] into n subintervals of width Dx and choose xi* from each interval. Then òf(x)dx=limåf(xi*)Dx.

a

n®¥

i=1

b

¥

is a function, F(x), such that F¢(x)=f(x). Indefinite Integral :òf(x)dx=F(x)+c where F(x) is an anti-derivative of f(x).

Fundamental Theorem of Calculus

Variants of Part I : Part I : If f(x) is continuous on [a,b] then

du(x)x

f(t)dt=u¢(x)féu(x)ùëûg(x)=òf(t)dt is also continuous on [a,b] òadxa

dbdx

f(t)dt=-v¢(x)féand g¢(x)=ftdt=fx. ()()ëv(x)ùû òvx()òdxadx

du(x)Part II : f(x)is continuous on[a,b], F(x) is f(t)dt=u¢(x)f[u(x)]-v¢(x)f[v(x)] òvx()dxan anti-derivative off(x)(i.e. F(x)=òf(x)dx) thenòf(x)dx=F(b)-F(a).

ab

òf(x)±g(x)dx=òf(x)dx±òg(x)dx bbb

òaf(x)±g(x)dx=òaf(x)dx±òag(x)dx òaòa

a

Properties

òcf(x)dx=còf(x)dx, c is a constant

òacf(x)dx=còaf(x)dx, c is a constant òa

b

bb

f(x)dx=0

f(x)dx=-òf(x)dx

b

b

a

a

b

f(x)dx=òf(t)dt

a

b

ba

ò

b

ba

f(x)dx£òf(x)dx

a

b

If f(x)³g(x) ona£x£bthen òf(x)dx³òg(x)dx Iff(x)³0 on a£x£b then òf(x)dx³0

a

If m£f(x)£Mon a£x£b then m(b-a)£òf(x)dx£M(b-a)

a

b

òkdx=kx+c nn1

òxdx=n+1x+c,n¹-1 òxdx=òxdx=lnx+c òax+bdx=alnax+b+c òlnudu=uln(u)-u+c òedu=e+c

+1

-11

11

u

u

Common Integrals òcosudu=sinu+c

òsinudu=-cosu+c òsecudu=tanu+c òsecutanudu=secu+c òcscucotudu=-cscu+c òcscudu=-cotu+c

22

òtanudu=lnsecu+c

òsecudu=lnsecu+tanu+c

u

du=tan(aa)+c òa+u

u1du=sin(a)+còa-u1

221

-1

-1

22Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

Calculus Cheat Sheet

Standard Integration Techniques

Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.

u Substitution : The substitution u=g(x)will convert òEx. ò5x2cos(x3)dx 12ba

f(g(x))g¢(x)dx=ò

5cos138538g(b)g(a)f(u)du using

du=g¢(x)dx. For indefinite integrals drop the limits of integration.

ò1213325x2cos(x3)dx=ò(u)duu=x3Þdu=3xdxÞxdx=du 32x=1Þu=1=1::x=2Þu=2=8 Integration by Parts : òudv=uv-òvdu and òudv=uv

ab

b

=53sin(u)1=b

(sin(8)-sin(1)) -òvdu. Choose u and dv from a

a

integral and compute du by differentiating u and compute v using v=òdv. Ex. òxe-xdx u=x

dv=e

-x

Þdu=dxv=-e

-x

Ex. òlnxdx 35-x-x-x-x-x

xedx=-xe+edx=-xe-e+c òò

u=lnx5dv=dxÞdu=1xdxv=x 55ò3lnxdx=xlnx3-ò3dx=(xln(x)-x)53=5ln(5)-3ln(3)-2

Products and (some) Quotients of Trig Functions For òsinnxcosmxdx we have the following : Foròtannxsecmxdx we have the following :

1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and

convert the rest to secants using cosines using sin2x=1-cos2x, then use

tan2x=sec2x-1, then use the substitution the substitution u=cosx.

u=secx. 2. m odd. Strip 1 cosine out and convert rest

2. m even. Strip 2 secants out and convert rest to sines using cos2x=1-sin2x, then use

to tangents using sec2x=1+tan2x, then the substitution u=sinx.

3. n and m both odd. Use either 1. or 2. use the substitution u=tanx. 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2.

and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be integral into a form that can be integrated. dealt with differently.

21

Trig Formulas : sin(2x)=2sin(x)cos(x), cos2(x)=12(1+cos(2x)), sin(x)=2(1-cos(2x)) Ex. òtan3xsec5xdx Ex. òsin3xdx cosx245òtanxsecxdx=òtanxsecxtanxsecxdx=ò(sec2x-1)sec4xtanxsecxdx=ò(u2-1)u4du=secx-secx+c17157535ò sin5xcos3xdx=ò=ò=(u=secx)(sin2x)2sinxsin4xsinxdx=dxcos3xcos3x(1-cos2x)2sinxdxu=cosxcos3x2224-(1-u3)du=-1-2u3+uduuuò() òò221=12secx+2lncosx-2cosx+cVisit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

Calculus Cheat Sheet

Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions.

sinq a2-b2x2Þx=abcos2q=1-sin2q Ex. òx=16a

secq b2x2-a2Þx=btanq a2+b2x2Þx=a

bsec2q=1+tan2q

tan2q=sec2q-1

óõ16x24-9x2dx 232sinq3Þdx=cosqdq 9122cosqdq=()3òsin2qdq4sin2q(2cosq) Use Right Triangle Trig to go back to x’s. From 3xRecall x2=x. Because we have an indefinite substitution we have sinq=2 so, integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d need to compute q’s and remove absolute value bars based on that and, xif x³0ì4-9x2From this we see that qcot=. So, x=í 3xî-xif x<014-9x2dx=-+c 2òx24-9x2xIn this case we have 4-9x=2cosq. Partial Fractions : If integrating ò

P(x)Q(x)224-9x2=4-4sinq=4cosq=2cosq =ò12cscdq=-12cotq+c2dx where the degree of P(x) is smaller than the degree of

Q(x). Factor denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the decomposition according to the following table.

Factor in Q(x) Term in P.F.D Factor in Q(x) Term in P.F.D AkA1A2Ak++L+ ax+b 2k(ax+b) ax+b(ax+b)(ax+b)ax+bax+bx+c Ex. ò(x-1)(x2+4)dx 7x2+13x(x-1)(x2+4)2Ax+B ax2+bx+c(ax2+bx+c) 7x2+13xkAkx+BkA1x+B1+L+ k22ax+bx+c(ax+bx+c)=x-1A+Bx+Cx2+4=A(x2+4)+(Bx+C)(x-1)(x-1)(x2+4) ò7x2+13x(x-1)(x23x+1dx=+dx-1xò+4)x2+4=òx4-1+3xx2+4+16x2+4dx 3=4lnx-1+2ln(x2+4)+8tan-1(x2)Here is partial fraction form and recombined. Set numerators equal and collect like terms. 7x2+13x=(A+B)x2+(C-B)x+4A-C Set coefficients equal to get a system and solve to get constants. A+B=7C-B=134A-C=0 A=4B=3C=16 An alternate method that sometimes works to find constants. Start with setting numerators equal in previous example : 7x2+13x=A(x2+4)+(Bx+C)(x-1). Chose nice values of x and plug in. For example if x=1 we get 20=5A which gives A=4. This won’t always work easily.

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

b

Calculus Cheat Sheet

Applications of Integrals

Net Area : òf(x)dx represents the net area between f(x) and the

a

x-axis with area above x-axis positive and area below x-axis negative.

Area Between Curves : The general formulas for the two main cases for each are, y=f(x)ÞA=ò

ba

éupper functionùëû

-éëlower functionùûdx & x=f(y)ÞA=ò

dc

éright functionù

ëû

-éëleft functionùûdy

If the curves intersect then the area of each portion must be found individually. Here are some

sketches of a couple possible situations and formulas for a couple of possible cases.

A=ò

ba

f(x)-g(x)dx

A=òf(y)-g(y)dy

c

d

A=òf(x)-g(x)dx+ò

ac

bc

g(x)-f(x)dx

Volumes of Revolution : The two main formulas are V=òA(x)dx and V=òA(y)dy. Here is some general information about each method of computing and some examples.

Rings Cylinders

A=2p(radius)(width / height) A=p(outer radius)2-(inner radius)2

()Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl.

Horz. Axis usef(x), Vert. Axis usef(y), Horz. Axis usef(y), Vert. Axis usef(x), g(x),A(x) and dx. Ex. Axis : y=a>0

g(y),A(y) and dy. Ex. Axis : y=a£0

g(y),A(y) and dy. Ex. Axis : y=a>0

g(x),A(x) and dx. Ex. Axis : y=a£0

outer radius :a-f(x) outer radius:a+g(x) radius :a-y

inner radius : a-g(x) inner radius:a+f(x) width : f(y)-g(y)

radius :a+y width : f(y)-g(y)

These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the y=a£0 case with a=0. For vertical axis of rotation (x=a>0 and x=a£0) interchange x and y to get appropriate formulas.

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

Calculus Cheat Sheet

Work : If a force ofF(x)moves an object ina£x£b, the work done is W=òF(x)dx

ab

Average Function Value : The average value of f(x) on a£x£b isfavg=

b1b-aa

òf(x)dx

Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, L=òds

ab

SA=ò2pyds (rotate about x-axis)

a

b

SA=ò2pxds (rotate about y-axis)

a

b

where ds is dependent upon the form of the function being worked with as follows. ds=1+ds=

()1+()dydxdxdy2dx if y=f(x),a£x£b dy if x=f(y),a£y£b

ds=

()dxdt22+()dydt22dt if x=f(t),y=g(t),a£t£b

2ds=r+(drdq)dq if r=f(q),a£q£b

With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds. With parametric and polar you will always need to substitute.

Improper Integral

An improper integral is an integral with one or more infinite limits and/or discontinuous integrands. Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value. This is typically a Calc II topic.

Infinite Limit 1. 3.

ò

¥

a

f(x)dx=limòf(x)dx

t®¥c

a

t

2.

¥

c

ò¥

-

b

f(x)dx=lim

t®-¥

òf(x)dx

t

b

ò¥f(x)dx=ò¥f(x)dx+ò

--ba

¥

f(x)dx provided BOTH integrals are convergent.

b

Discontinuous Integrand

1. Discont. at a:òf(x)dx=limf(x)dx +ò

t®a

tb

c

a

a

2. Discont. at b :òf(x)dx=limf(x)dx-ò

a

t®b

a

bc

bt

3. Discontinuity at aComparison Test for Improper Integrals : If f(x)³g(x)³0on [a,¥) then, 1. Ifòf(x)dx conv. thenòg(x)dx conv.

a

a

¥

¥

2. Ifòg(x)dx divg. thenòf(x)dx divg.

a

a

¥¥

Useful fact : If a>0 then ò

¥a

x

1pdx converges if p>1 and diverges for p£1.

Approximating Definite Integrals

a

For given integral òf(x)dx and a n (must be even for Simpson’s Rule) define Dx=b-n and

ab

divide [a,b] into n subintervals [x0,x1], [x1,x2], … , [xn-1,xn] with x0=a and xn=b then,

****ùMidpoint Rule : òf(x)dx»Dxéfx+fx+L+fx, x()()()12nûi is midpoint [xi-1,xi] ëa

bDx

Trapezoid Rule : òf(x)dx»éf(x0)+2f(x1)++2f(x2)+L+2f(xn-1)+f(xn)ùû a2ë

bDx

Simpson’s Rule : òf(x)dx»éf(x0)+4f(x1)+2f(x2)+L+2f(xn-2)+4f(xn-1)+f(xn)ù ëûa3

b

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins