BC Calc Memory Quiz

lim (x→c⁻) f(x)=lim(x→c⁺)f(x)=f(c) OR lim(x→c)f(x)=f(c)

(a) since f is continuous on [a,b] and

(b) f(a)<k<f(b),

(c) IVT guarantees at least one c in (a,b) such that f(c)=k

(d/dx)(f(g(x)))=f’(g(x))*g’(x)

(d/dx)[f*g]= f’g+fg’

(d/dx)[f/g]= (f’g-fg’)/g²

cosu(u’)

-sinu(u’)

sec²u(u’)

-csc²u(u’)

secutanu(u’)

-cscucotu(u’)

since f is a continuous function on [a,b] must have a maximum and minimum value within that interval.

a high/low point relative to the points around it; can only occur at a critical value.

the highest/lowest point on a given interval; can occur at a critical value OR an endpoint.

Since f is continuous on [a, b] and differentiable on (a, b) MVT guarantees that there exists an x-value such that f’(x)=(f(b)-f(a))/(b-a)

sin u +C

-cos u +C

tan u +C

-cot u +C

sec u +C

-csc u +C

∫(a to b) ƒ’(x)dx=f(b)-f(a)

(d/dx)∫(u to v) g(t)dt=g(v)(v’)-g(u)(u’)

(1/(b-a))∫(a to b) f(x)dx

∫(a to b) v(t)dt

∫(a to b) |v(t)| dt

-ln|cos u|+C

ln|sin u| +C

ln|secu+tanu|+C

-ln|cscu+cotu|+C

(f⁻¹)’(x)=1/(f’(f⁻¹(x))

eⁿ+c

aⁿ(ln a) n’

u’/u

u’/(u (ln n))

eⁿ+C

aⁿ/lna +C

ln |u| +C

u’/√(1-u²)

u’/(1+u²)

u’/(u√(u²-1))

-u’/√(1-u²)

-u’/(1+u²)

-u’/(u√(u²-1))

arcsin (u/a) +C

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

if dy/dt=ky, then y=c(e^kt)

if dy/dt =ky(1-(y/L)), then y=L/(1+C(e^-kt))

∫(a to b) (top-bottom)dx OR ∫(a to b) (right-left)dy

v=π(∫(a to b) R²dx

v=π(∫(a to b) (R²-r²)dx

v=2π(∫(a to b) (ph)dx

v=(∫(a to b) A(x)dx

L=∫(a to b) √(1+[f’(x)]²)dx

S=∫(a to b) 2πr√(1+[f’(x)]²)dx

∫u dv = uv-∫v du

f(x) = f(c)+f’(c)(x-c)+f’’(c)(x-c)²/2!+…+fⁿ(c)(n-c)ⁿ/n!

1+x+x²/2!+x³/3!+…+xⁿ/n! (for all real numbers)

x-x³/3!+x⁵/5!-x⁷/7!+…+(-1)ⁿx²ⁿ⁺¹/(2n+1)! (for all real numbers)

1-x²/2!+x⁴/4!-x⁶/6!+…+(-1)ⁿx²ⁿ/(2n)! (for all real numbers)

1+x+x²+x³+x⁴+…+xⁿ (for -1<x<1)

error≤|Aₙ₊₁|

error≤|fⁿ⁺¹(max)|/(n+1)!×(x-c)ⁿ

dy/dx= (dy/dt)/(dx/dt)

d²y/dx²=(d/dt)[dy/dx]/(dx/dt)

√[(dx/dt)²+(dy/dt)²]

L=∫(a to b) √[(dx/dt)²+(dy/dt)²] dt

A=½∫(a to b) r²dθ

x=r cosθ and y=r sinθ