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Midpoint rule

b

a

f(x) d x

n

i=1

f( xi) x = x [f(x1)+ + f(xn)]

where (xi1,xi) and

x =b a

n and

xi=1

2(xi1+xi)

Trapezoidal rule

b

a

f(x) d x Tn=x

2 [f(x0) + 2f(x1) + 2f(x2) + + 2f(xn1) + f(xn)]

where x =b a

n and

xi= a +ix

Midpoint and trapezoidal error bounds

ETand EMare the errors in the trapezoidal and midpoint rules

ET K(b a)3

12n2

and

EM K(b a)3

24n2

where fn(x) K

for ax b

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Simpsons rule

b

a

f(x) d x Sn=x

3 [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + + 2f(xn2) + 4f(xn1) + f(xn)]

where nis even and

x =b a

n

Simpsons error bounds

ESis the error in Simpsons rule

ES K(b a)5

180n4

where f(4)(x) K

for

ax b

Symmetric functions

Suppose fis continuous on [a, a].

If fis even [f(x) =f(x)], then a

a

f(x) d x = 2a

0

f(x) d x

If fis odd [f(x) = f(x)], then

a

a

f(x) d x = 0

Net change theorem

b

a

F(x) d x =F(b) F(a)

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Fundamental theorem of calculus

Suppose fis continuous on [a, b].

Part 1

Given integral How to solve it

f(x) = x

a

f(t) dt

Plug xin for t.

f(x) = a

x

f(t) dt Reverse limits of integration and multiply by 1, then

plug xin for t.

f(x) = g(x)

a

f(t) dt Plug g(x)in for t, then multiply by dg/d x.

f(x) = a

g(x)

f(t) dt Reverse limits of integration and multiply by 1, then

plug g(x)in for tand multiply by dg/d x.

f(x) =

h(x)

g(x)f(t) dt

Split the limits of integration as

0

g(x)f(t) dt+

h(x)

0f(t) dt.

Reverse limits of integration on 0

g(x)

f(t) dtand multiply

by 1, then plug g(x)and h(x)in for t, multiplying by

dg/d xand dh /d xrespectively.

Part 2

b

a

f(x) d x =F(b) F(a)

where Fis any antiderivative of f, that is, a function such that F =f

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Integration by parts

u dv =u v v du

Properties of integrals

b

a

c d x =c(b a)

b

a

f(x) + g(x) d x = b

a

f(x) d x+ b

a

g(x) d x

b

a

cf(x) d x = c

b

a

f(x) d x

b

a

f(x) g(x) d x = b

a

f(x) d x b

a

g(x) d x

Common indefinite integrals

k d x = k x+ C

xn d x =xn+1

n+ 1+C with n 1

1

xd x = ln x +C

ex d x =ex +C ax d x =ax

ln a+C

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Integrals of trig functions

sinx d x = cosx+ C cscx d x = ln cscx cotx +C

or cscx d x = ln(sinx

2) ln(cosx

2) +C

cosx d x = sinx + C secx d x = ln secx+ tanx +C

or secx d x = ln(sinx

2+ cos

x

2) ln(cosx

2 sin

x

2) +C

tanx d x = ln cosx+ C

cotx d x = ln sinx + C

Other common trig integrals

sec2 x d x = tanx + C csc2 x d x = cotx + C

secxtanx d x = secx+ C

cscxcotx d x = cscx+ C

1

x2 + 1d x = tan1x+ C

1

1 x2d x = sin1x+ C

sinhx d x = coshx + C coshx d x = sinhx + C

Integrals of inverse hyperbolic trig functions

sinh1x d x = xsinh1x x2 + 1 +C

cosh1x d x = xcosh1x x 1 x+ 1 +C

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tanh1x d x =1

2log(1 x2) + x tanh1x+ C

coth1x d x =1

2log(1 x2) + xcoth1x+ C

Integrals resulting in inverse hyperbolic trig functions

1

x2 + 1d x = sinh1x

1

x 1 x+ 1d x = cosh1x

1

1 x2d x = tanh1x

1

1 x2d x = coth1x

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Trig substitution setup

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