Types of Integration (By parts, U-Sub, partial)
int (ln x)^2/(x)dx
(ln x)^3/3 + C
sum_(n=2)^oo (7/(n(n+1)))
Will the series be convergent or divergent? How will you prove that both series using the Comparison Test are convergent or divergent?
sum_(n=2)^oo 7/n^2
will converge by the p-series Test (p=2>1)
If bn, which is
sum_(n=2)^oo 7/n^2
is convergent then that means an, which is
sum_(n=2)^oo 7/(n(n+1))
is convergent.
int (sin^3x)(cos^2x)dx
((-cos^3)/x+(cos^5x)/x)+C
Without writing it down, tell me if this is convergent or divergent, and why?
sum_(n=1)^oo 1/n
It's Divergent because it's a harmonic series
What conditions have to be satisfied to use the Integral Test?
What conditions does an alternating series converge?
f must be positive, continuous, and decreasing for x>=1, and
a_n = f(n)
The alternating series
sum_(n=1)^oo (-1)^(n-1) b_n
converges if
I)
b_(n+1) <= b_n
II)
lim_(n->oo) b_n=0
sum_(n=1)^oo (n/3^n)
sum_(n=1)^oo (n/3^n) -> lim_(n->oo)(n^(1/n)/3)=1/3
The series converges.
Double Points
Integrate:
int xsqrt(4-x)dx
a)by parts, letting dv =
sqrt(4-x)dx
b) by substitution, letting u= 4-x
A)
(-2/3x(4-x)^(3/2)-4/15(4-x)^(5/2)+C)
B)
(2/5(4-x)^(5/2)-8/3(4-x)^(3/2)+C)
sum_(n=1)^oo (n-1)/(sqrt(n^6+1)
The series is convergent by a comparison test with the convergent p-series
sum_(n=2)^oo 1/n^2
int (cos^5t)/(sqrt(sint)
2*(sint)^(1/2)-4/5(sint)^(5/2)+2/9(sint)^(9/2)+C
sum_(n=1)^oo 1/n^(3/2)
Convergent p-series with p = 3/2 > 1
sum_(n=1)^oo (ln n)/n^2
int_1^oo (ln x)/x^2 dx = [-(ln x+1)/x]_1^oo = 1; converges
sum_(n=1)^oo (9^n/n^5)
sum_(n=1)^oo (9^n/n^5) -> lim_(n->oo) ((9n^5)/(n+1)^5)=9>1
This series is divergent
int(x^3/(2+x^4)^2)dx
-(1/4(2+x^4))+C
sum_(n=1)^oo (4n^2-n)/(n^3+9)
lim_(n->oo) [(4n^3-n^2)/(4n^3+36)] =1
Diverge
intsec^2(x)sec^2(x)sqrt(tanx)dx
2/3tan^(3/2)(x)+2/7tan^(7/2)(x)+C
sum_(n=0^oo 5(-4/3)^n
Use Geometric Series Test
By the Geometric Test it diverges
sum_(n=1)^oo n/(n^4+2n^2+1)
Double Points: Identify the two series that are the same and tell me if they are convergent or divergent.
A)
sum_(n=1)^oo (n5^n)/(n!)
B)
sum_(n=0)^oo (n5^n)/((n+1)!)
C)
sum_(n=0)^oo ((n+1)5^(n+1))/((n+1)!)
(a) and (c) are the same
sum_(n=1)^oo (n5^n)/(n!) = sum_(n=0)^oo ((n+1)5^(n+1))/((n+1)!)
Both series converge.
int (x-4)/(x^2+2x-15)dx
9/8 ln|x+5| - 1/8 ln|x-3| + C
Use the Limit Comparison Test to determine the convergence or divergence of the series.
sum_(n=1)^oo (1/(nsqrt(n^2+1)))
sum_(n=1)^oo (1/(nsqrt(n^2+1)))
converges by a limit comparison with the convergent p-series
sum_(n=1)^oo 1/n^2
int(x^3)/(sqrt(16-x^2))dx
64(-1/4sqrt(16-x^2)+1/192(16-x^2)^(3/2)+C)
sum_(n=1)^oo (cos n+2)/(sqrt(n))
Use the Direct Comparison Test to determine the convergence or divergence of the series.
P-series Test:
(1/sqrtn)=1/n^(1/2); p=1/2, 0<p<=1
which means that the series is divergent.
Use Alternating Series Test
sum_(n=1)^oo ((-1)^(n+1)/5^n)
Convergent
lim_(n->oo) 1/5^n=1/oo = 0
sum_(n=1)^oo ((n)/500)^n
By the Ratio Test the series diverges
int (x^2+4)/(3x^3+4x^2-4x) dx
Use partial fractions to solve this equation.
(x-2)/(x(3x-2))=1/x-2/(3x-2)
Determine whether the series converges or diverges
sum_(n>=1)^oo (2n^2+3n)/(sqrt(5+n^5)
Use the L.C.T with the sereis
sum_(n>=1)^oo (2n^2)/sqrt(n^5)
Thus, by the L.C.T., the series in question is of the same nature as the series
sum_(n=1)^oo (2)/(n^(1/2)
intx/(sqrt(4x-x^2))dx
Find the Indefinite Integral
2sin^(-1)*(x-2)/2-sqrt(4x-x^2)+C
sum_(n=0)^oo 2^n * 1/(3^(n-1))
Convergent or Divergent?
Convergent
Determine whether the series converges absolutely or conditionally
sum_(n=1)^oo (cosnpi)/(n+1)
The given series converges by the A.S.T., but
sum_(n=0)^oo |cosnpi|/(n+1) = sum_(n=0)^oo 1/(n+1)
diverges by a limit comparison to the divergent harmonic series
sum_(n=1)^oo 1/n*lim_(n->oo) (|cosnpi|/(n+1))/(1/n)=1
So the series converges conditionally.
sum_(n=1)^oo |((-1)^n3^(n+2))/((n-1)^n)|
Use the Root Test to find the answer
This series is convergent.
Go here for the step-by-step:
https://www.youtube.com/watch?v=ahf0eXOll1M&ab_channel=TheOrganicChemistryTutor