Studied by 24 people∫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
choose u
find du
rewrite integral in terms of u
evaluate integral
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
Note
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