AP Calc-integrals

∫du

∫du

u + C

∫edu

e^u + C

∫cos(u)du

sin(u) + C

∫cot(u)du

lnIsin(u)I + C

∫csc(u)du

-lnIcsc(u) + cot(u)I + C

∫csc²(u)du

-cot(u) + C

∫csc(u)cot(u)du

-csc(u) + C

∫du/(a²+u²)

(1/a)arctan(u/a) + C

∫[f(u) + g(u)]du

∫f(u)du + ∫g(u)du

∫[f(u) - g(u)]du

∫f(u)du - ∫g(u)du

∫(a^u)du

(1/ln(a))a^u +C

∫sin(u)du

-cos(u) + C

∫tan(u)du

-lnIcos(u)I + C

∫sec(u)du

lnIsec(u) + tan(u)I + C

∫sec²(u)du

tan(u) + C

∫sec(u)tan(u)du

sec(u) + C

∫du/√(a²-u²)

arcsin(u/a) +C

∫du/[(u)√(u²−a²)

(1/a)arcsec(IuI/a) + C

net change formula

∫f'(x)=f(b) - f(a)

when do you use net change formula?

when the question asks for a rate of change

∆x=?

(b-a)/n

xi=

a + i∆x

area under the curve=

∆x(∑heights)

trapezoidal rule

(b-a)/(2n)[f(a) + 2f(n-1) + f(b)]

how do you do a midpoint sum

you take the ∆x of the points and multiply it by the middle number

∑c=

cn

∑

n(n+2)/2

∑i²=

n(n+1)(2n+1)/6

∑i³=

n²(n+1)²/4

∑(ai + bi)=

∑ai + ∑bi

∫f(x)dx=

limn→∞∑f(xi)∆x

(d/dx)g(x)=∫(from x to 0)√1+t² dt(d/dx)=

g'(x)=√1+x² (1) - √1+0² (0)

∫x^n=

(x^n+1)/(n+1)

∫(from a to a)f(x)dx=

0

∫(from b to a)f(x)dx=

-∫(from a to b)f(x)dx

f(avg)=

1/(b-a)∫(from a to b)f(x)dx

steps for u-sub

1. choose u

2. find du

3. rewrite integral in terms of u

4. evaluate integral

5. replace u with function

if f is even, f(-x)=f(x), then ∫(from -a to a)f(x)dx=

2∫(from 0 to a)f(x)dx

if f is odd, f(-x)=-f(x), then ∫(from -a to a)f(x)dx=

0

when n/d, n is 2 less, use...

inverse trig

when n/d, n=d, use...

u-sub

when n/d, n≥d, use...

long division