I'm going to exploit the use of LaTeX for now. As soon as I can program the AcadaWiki , I will stop.
edit
1. True or False
∫
x
2
d
x
=
1
3
x
3
{\displaystyle \int x^{2}\,dx={\frac {1}{3}}x^{3}}
2. True or False
An antiderivative of
x
2
is
1
3
x
3
{\displaystyle {\mbox{An antiderivative of }}x^{2}{\mbox{ is }}{\frac {1}{3}}x^{3}}
3. True or False
∫
cos
x
d
x
=
sin
x
+
c
−
10
{\displaystyle \int \cos x\,dx=\sin x+c-10}
4. Evaluate
∫
{
x
3
−
1
x
2
}
d
x
{\displaystyle \int \left\{{\frac {x^{3}-1}{x^{2}}}\right\}\,dx}
A.
x
5
−
x
2
B.
1
3
(
x
3
−
1
)
2
C.
1
2
x
2
+
1
x
D.
1
2
x
2
+
1
x
+
c
E.
1
3
(
x
3
−
1
)
2
+
c
{\displaystyle {\begin{aligned}&{\mbox{A. }}x^{5}-x^{2}\\&{\mbox{B. }}{\frac {1}{3}}\,(x^{3}-1)^{2}\\&{\mbox{C. }}{\frac {1}{2}}\,x^{2}+{\frac {1}{x}}\\&\color {red}{\mbox{D. }}{\frac {1}{2}}\,x^{2}+{\frac {1}{x}}+c\\&{\mbox{E. }}{\frac {1}{3}}(x^{3}-1)^{2}+c\\\end{aligned}}}
5. Evaluate
∫
{
(
5
x
2
+
1
)
(
5
x
3
+
3
x
−
8
)
6
}
d
x
{\displaystyle \int {\Big \{}(5x^{2}+1)(5x^{3}+3x-8)^{6}{\Big \}}\,dx}
A.
1
21
(
5
x
3
+
3
x
−
8
)
7
{\displaystyle {\mbox{A. }}{\frac {1}{21}}(5x^{3}+3x-8)^{7}}
B.
1
21
(
5
x
3
+
3
x
−
8
)
7
+
c
{\displaystyle {\mbox{B. }}{\frac {1}{21}}(5x^{3}+3x-8)^{7}+c}
C.
(
5
x
2
+
1
)
(
5
x
3
+
3
x
−
8
)
6
+
c
{\displaystyle {\mbox{C. }}(5x^{2}+1)(5x^{3}+3x-8)^{6}+c}
D.
1
2
x
2
+
1
x
+
c
{\displaystyle \color {red}{\mbox{D. }}{\frac {1}{2}}\,x^{2}+{\frac {1}{x}}+c}
E.
1
3
(
x
3
−
1
)
2
+
c
{\displaystyle {\mbox{E. }}{\frac {1}{3}}(x^{3}-1)^{2}+c}
6. True or False
∫
x
n
d
x
=
1
n
+
1
x
n
+
1
+
c
if
n
≠
−
1
{\displaystyle \int x^{n}\,dx={\frac {1}{n+1}}x^{n+1}+c\,{\mbox{ if }}n\neq -1}
7. True or False
∫
[
f
(
x
)
]
n
d
x
=
1
n
+
1
[
f
(
x
)
]
n
+
1
+
c
if
n
≠
−
1
{\displaystyle \int {\Big [}f(x){\Big ]}^{n}\,dx={\frac {1}{n+1}}{\Big [}f(x){\Big ]}^{n+1}+c\,{\mbox{ if }}n\neq -1}
8. True or False
∫
[
f
(
x
)
]
n
f
′
(
x
)
d
x
=
1
n
+
1
[
f
(
x
)
]
n
+
1
+
c
if
n
≠
−
1
{\displaystyle \int {\Big [}f(x){\Big ]}^{n}\,f^{\prime }(x)\,dx={\frac {1}{n+1}}{\Big [}f(x){\Big ]}^{n+1}+c\,{\mbox{ if }}n\neq -1}
9. True or False
∫
(
x
+
7
)
50
d
x
=
1
51
(
x
+
7
)
51
+
c
{\displaystyle \int (x+7)^{50}\,dx={\frac {1}{51}}(x+7)^{51}+c}
10. True or False
∫
(
2
x
+
7
)
50
d
x
=
1
51
(
2
x
+
7
)
51
+
c
{\displaystyle \int (2x+7)^{50}\,dx={\frac {1}{51}}(2x+7)^{51}+c}
11. True or False
∫
d
x
=
x
−
c
{\displaystyle \int \,dx=x-c}
12. True or False
∫
[
f
(
x
)
]
2
d
x
=
1
3
[
f
(
x
)
]
3
+
c
{\displaystyle \int {\Big [}f(x){\Big ]}^{2}\,dx={\frac {1}{3}}{\Big [}f(x){\Big ]}^{3}+c}
13. Evaluate
∫
(
x
(
3
x
2
+
1
)
4
)
d
x
{\displaystyle \int {\Big (}x(3x^{2}+1)^{4}{\Big )}\,dx}
A.
1
5
(
3
x
2
+
1
)
5
+
c
{\displaystyle {\mbox{A. }}{\frac {1}{5}}(3x^{2}+1)^{5}+c}
B.
x
5
(
3
x
2
+
1
)
5
+
c
{\displaystyle {\mbox{B. }}{\frac {x}{5}}(3x^{2}+1)^{5}+c}
C.
1
30
(
3
x
2
+
1
)
5
+
c
{\displaystyle \color {Red}{\mbox{C. }}{\frac {1}{30}}(3x^{2}+1)^{5}+c}
D.
x
5
(
3
x
2
+
1
)
5
{\displaystyle {\mbox{D. }}{\frac {x}{5}}(3x^{2}+1)^{5}}
E.
1
5
(
3
x
2
+
1
)
5
{\displaystyle {\mbox{E. }}{\frac {1}{5}}(3x^{2}+1)^{5}}
14. Evaluate
∫
(
x
2
cos
(
x
3
)
)
d
x
{\displaystyle \int {\Big (}x^{2}\cos(x^{3}){\Big )}\,dx}
A.
x
2
cos
x
3
+
c
{\displaystyle \color {black}{\mbox{A. }}x^{2}\cos {x^{3}}+c}
B.
−
1
3
sin
x
3
3
x
3
+
c
{\displaystyle {\mbox{B. }}-{\frac {1}{3}}\sin {x^{3}}3x^{3}+c}
C.
1
3
sin
x
3
+
c
{\displaystyle {\mbox{C. }}{\frac {1}{3}}\sin {x^{3}}+c}
D.
1
3
sin
x
3
3
x
2
+
c
{\displaystyle {\mbox{D. }}{\frac {1}{3}}\sin {x^{3}}3x^{2}+c}
E.
cos
3
x
2
+
c
{\displaystyle \color {black}{\mbox{E. }}\cos {3x^{2}}+c}
Evaluation of Multiple Derivatives
edit
Dr. Euler assigned a recent extra credit assignment involving the finding of an equation to find the multiple derivatives of a function . The below equations are what I produced, yet they remain to be proved mathematically.
For a monomial to the power of a fraction, or to the power of a negative number
{\displaystyle \color {Black}{\mbox{For a monomial to the power of a fraction, or to the power of a negative number}}}
If
f
(
x
)
=
a
x
n
1
n
2
, then
f
(
m
)
(
x
)
=
a
x
(
n
1
n
2
−
m
)
∗
[
∏
h
=
1
m
{
n
1
−
n
2
(
m
−
h
)
}
]
n
2
m
with
m
∈
N
and
a
≠
0
{\displaystyle {\mbox{If }}f(x)=ax^{\frac {n_{1}}{n_{2}}}{\mbox{, then }}f^{(m)}(x)={\frac {ax^{{\big (}{\frac {n_{1}}{n_{2}}}-m{\big )}}*{\Bigg [}\prod _{h=1}^{m}{\Big \{}n_{1}-n_{2}(m-h){\Big \}}{\Bigg ]}}{n_{2}^{m}}}{\mbox{ with }}m\in \mathbb {N} {\mbox{ and }}a\neq 0}
For a monomial to the power of a natural number
{\displaystyle \color {Black}{\mbox{For a monomial to the power of a natural number}}}
If
f
(
x
)
=
a
x
n
, then
f
(
m
)
(
x
)
=
[
n
P
m
]
a
x
(
n
−
m
)
with
n
∈
N
and
m
∈
N
and
a
≠
0
{\displaystyle {\mbox{If }}f(x)=ax^{n}{\mbox{, then }}f^{(m)}(x)={\Big [}{}_{n}P_{m}{\Big ]}ax^{(n-m)}{\mbox{ with }}n\in \mathbb {N} {\mbox{ and }}m\in \mathbb {N} {\mbox{ and }}a\neq 0}
Dr. Euler's approach follows a different method, using limits as a brute-force method. Let us find multiple derivatives of the function
f
(
x
)
{\displaystyle f(x)}
f
′
(
x
)
=
lim
h
→
0
[
f
(
x
+
h
)
−
f
(
x
)
h
]
{\displaystyle f^{\prime }(x)=\lim _{h\to 0}{\Bigg [}{\frac {f(x+h)-f(x)}{h}}{\Bigg ]}}
f
′
′
(
x
)
=
lim
h
→
0
[
f
′
(
x
+
h
)
−
f
′
(
x
)
h
]
=
lim
h
→
0
[
f
(
{
x
+
h
}
+
h
)
−
f
(
x
+
h
)
h
−
f
(
x
+
h
)
−
f
(
x
)
h
h
]
=
lim
h
→
0
[
f
(
x
+
2
h
)
−
f
(
x
+
h
)
−
f
(
x
+
h
)
+
f
(
x
)
h
h
]
=
lim
h
→
0
[
f
(
x
+
2
h
)
−
2
f
(
x
+
h
)
+
f
(
x
)
h
2
]
{\displaystyle {\begin{alignedat}{4}f^{\prime \prime }(x)&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime }(x+h)-f^{\prime }(x)}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {{\frac {f(\{x+h\}+h)-f(x+h)}{h}}-{\frac {f(x+h)-f(x)}{h}}}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {\frac {f(x+2h)-f(x+h)-f(x+h)+f(x)}{h}}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f(x+2h)-2f(x+h)+f(x)}{h^{2}}}{\Bigg ]}\\\end{alignedat}}}
f
′
′
′
(
x
)
=
lim
h
→
0
[
f
′
′
(
x
+
h
)
−
f
′
′
(
x
)
h
]
=
lim
h
→
0
[
f
′
(
x
+
2
h
)
−
2
f
′
(
x
+
h
)
+
f
′
(
x
)
h
2
]
=
lim
h
→
0
[
f
(
{
x
+
2
h
}
+
h
)
−
f
(
x
+
2
h
)
h
−
2
f
(
{
x
+
h
}
+
h
)
−
f
(
x
+
h
)
h
+
f
(
x
+
h
)
−
f
(
x
)
h
h
2
]
=
lim
h
→
0
[
f
(
x
+
3
h
)
−
f
(
x
+
2
h
)
−
2
f
(
x
+
2
h
)
+
2
f
(
x
+
h
)
+
f
(
x
)
h
)
−
f
(
x
)
h
h
2
]
=
lim
h
→
0
[
f
(
x
+
3
h
)
−
3
f
(
x
+
2
h
)
+
3
f
(
x
+
h
)
−
f
(
x
)
h
3
]
{\displaystyle {\begin{alignedat}{6}f^{\prime \prime \prime }(x)&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime \prime }(x+h)-f^{\prime \prime }(x)}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime }(x+2h)-2f^{\prime }(x+h)+f^{\prime }(x)}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {{\frac {f(\{x+2h\}+h)-f(x+2h)}{h}}-2{\frac {f(\{x+h\}+h)-f(x+h)}{h}}+{\frac {f(x+h)-f(x)}{h}}}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {\frac {f(x+3h)-f(x+2h)-2f(x+2h)+2f(x+h)+f(x)h)-f(x)}{h}}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}}{\Bigg ]}\\\end{alignedat}}}
This was the extent to Dr. Euler's illustration of taking multiple derivatives. Obviously Dr. Euler's method trumps over mine because it has already been proven to solve any function, including trigonometric functions and rational functions . Upon analyzing these formulas, I formulated an equation for finding the nth derivative of a function with the help of Isaac Nichols, one of my fellow students.f(x) . The derivatives of of this function are as follow:
Derivative
Prime Notation
Limit notation
Binomial coefficients
0
f
(
x
)
{\displaystyle \color {Black}f(x)}
lim
h
→
0
f
(
x
)
{\displaystyle \lim _{h\to 0}f(x)}
1
1
s
t
{\displaystyle 1^{st}}
f
′
(
x
)
{\displaystyle f^{\prime }(x)}
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
{\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}
1 1
2
n
d
{\displaystyle 2^{nd}}
f
′
′
(
x
)
{\displaystyle f^{\prime \prime }(x)}
lim
h
→
0
[
f
(
x
+
2
h
)
−
2
f
(
x
+
h
)
+
f
(
x
)
h
2
]
{\displaystyle \lim _{h\to 0}{\Bigg [}{\frac {f(x+2h)-2f(x+h)+f(x)}{h^{2}}}{\Bigg ]}}
1 2 1
3
r
d
{\displaystyle 3^{rd}}
f
′
′
′
(
x
)
{\displaystyle f^{\prime \prime \prime }(x)}
lim
h
→
0
[
f
(
x
+
3
h
)
−
3
f
(
x
+
2
h
)
+
3
f
(
x
+
h
)
−
f
(
x
)
h
3
]
{\displaystyle \lim _{h\to 0}{\Bigg [}{\frac {f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}}{\Bigg ]}}
1 3 3 1
⋮
{\displaystyle \vdots }
⋮
{\displaystyle \vdots }
⋮
{\displaystyle \vdots }
⋮
{\displaystyle \vdots }
n
t
h
{\displaystyle n^{th}}
f
(
n
)
(
x
)
{\displaystyle \color {black}f^{(n)}(x)}
lim
h
→
0
[
∑
i
=
0
n
[
(
−
1
)
i
(
n
i
)
f
(
x
+
{
n
−
i
}
h
)
]
h
n
]
{\displaystyle \lim _{h\to 0}{\Bigg [}{\frac {\sum _{i=0}^{n}{\Big [}(-1)^{i}{\binom {n}{i}}f(x+\{n-i\}h){\Big ]}}{h^{n}}}{\Bigg ]}}
1 n
(
n
2
)
…
(
n
n
−
2
)
{\displaystyle {\binom {n}{2}}\,\dots \,{\binom {n}{n-2}}}
Calculus II Equations
edit
A
=
∫
a
b
f
(
x
)
d
x
for individual curves.
=
∫
a
b
[
f
(
x
)
−
g
(
x
)
]
d
x
for multiple curves.
=
∫
c
d
[
h
(
y
)
−
k
(
y
)
]
d
y
for multiple curves in respect of the y axis.
V
=
π
∫
a
b
f
(
x
)
d
x
for the Disk Method.
=
π
∫
a
b
[
f
2
(
x
)
−
g
2
(
x
)
]
d
x
for the Washer Method.
=
∫
a
b
A
(
x
)
d
x
with
A
(
x
)
=
the area of cross section.
=
2
π
∫
a
b
[
x
f
2
(
x
)
]
d
x
for Shell Method rotation via the y axis.
=
2
π
∫
a
b
x
[
f
(
x
)
−
g
(
x
)
]
d
x
for Shell Method rotation via the y axis between two curves.
=
π
∫
a
b
[
(
outer r
)
2
−
(
inner r
)
2
]
d
x
for rotation around an alternative axis.
L
=
∫
a
b
1
+
[
f
′
(
x
)
]
2
d
x
for Curve Length.
=
∫
a
b
[
f
′
(
t
)
]
2
+
[
g
′
(
t
)
]
2
d
t
with Parametric Equations.
{\displaystyle {\begin{aligned}A&=\int _{a}^{b}f(x)\,dx{\mbox{ for individual curves.}}\\&=\int _{a}^{b}{\big [}f(x)-g(x){\big ]}\,dx{\mbox{ for multiple curves.}}\\&=\int _{c}^{d}{\big [}h(y)-k(y){\big ]}\,dy{\mbox{ for multiple curves in respect of the y axis.}}\\V&=\pi \int _{a}^{b}f(x)\,dx{\mbox{ for the Disk Method.}}\\&=\pi \int _{a}^{b}{\big [}f^{2}(x)-g^{2}(x){\big ]}\,dx{\mbox{ for the Washer Method.}}\\&=\int _{a}^{b}A(x)\,dx{\mbox{ with }}A(x)={\mbox{ the area of cross section.}}\\&=2\pi \int _{a}^{b}{\big [}xf^{2}(x){\big ]}\,dx{\mbox{ for Shell Method rotation via the y axis.}}\\&=2\pi \int _{a}^{b}x{\big [}f(x)-g(x){\big ]}\,dx{\mbox{ for Shell Method rotation via the y axis between two curves.}}\\&=\pi \int _{a}^{b}{\big [}({\mbox{outer r}})^{2}-({\mbox{inner r}})^{2}{\big ]}\,dx{\mbox{ for rotation around an alternative axis.}}\\L&=\int _{a}^{b}{\sqrt {1+[f^{\prime }(x)]^{2}}}\,dx{\mbox{ for Curve Length.}}\\&=\int _{a}^{b}{\sqrt {[f^{\prime }(t)]^{2}+[g^{\prime }(t)]^{2}}}\,dt{\mbox{ with Parametric Equations.}}\\\end{aligned}}}
The Volume of a Cylindrical Wedge
edit
edit
![{\displaystyle \int {\Big [}f(x){\Big ]}^{n}\,dx={\frac {1}{n+1}}{\Big [}f(x){\Big ]}^{n+1}+c\,{\mbox{ if }}n\neq -1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28f388e1566d744d2b39ac56ac745892b5784b32)
![{\displaystyle \int {\Big [}f(x){\Big ]}^{n}\,f^{\prime }(x)\,dx={\frac {1}{n+1}}{\Big [}f(x){\Big ]}^{n+1}+c\,{\mbox{ if }}n\neq -1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26bd5936ef03da8c00890ba42e66e98228f5e4ff)
![{\displaystyle \int {\Big [}f(x){\Big ]}^{2}\,dx={\frac {1}{3}}{\Big [}f(x){\Big ]}^{3}+c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aaa4e4b2e5aea92cc8e640a82cb85edb0c5f6147)
![{\displaystyle {\mbox{If }}f(x)=ax^{\frac {n_{1}}{n_{2}}}{\mbox{, then }}f^{(m)}(x)={\frac {ax^{{\big (}{\frac {n_{1}}{n_{2}}}-m{\big )}}*{\Bigg [}\prod _{h=1}^{m}{\Big \{}n_{1}-n_{2}(m-h){\Big \}}{\Bigg ]}}{n_{2}^{m}}}{\mbox{ with }}m\in \mathbb {N} {\mbox{ and }}a\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a9df483ea9848685a33c0887e82a9840f7b780f)
![{\displaystyle {\mbox{If }}f(x)=ax^{n}{\mbox{, then }}f^{(m)}(x)={\Big [}{}_{n}P_{m}{\Big ]}ax^{(n-m)}{\mbox{ with }}n\in \mathbb {N} {\mbox{ and }}m\in \mathbb {N} {\mbox{ and }}a\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fafb2d6e4e095d7684f81b8be4cdfea5e39893b)
![{\displaystyle f^{\prime }(x)=\lim _{h\to 0}{\Bigg [}{\frac {f(x+h)-f(x)}{h}}{\Bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77625ddcf8bd48697272965a25d2a20264e1a195)
![{\displaystyle {\begin{alignedat}{4}f^{\prime \prime }(x)&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime }(x+h)-f^{\prime }(x)}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {{\frac {f(\{x+h\}+h)-f(x+h)}{h}}-{\frac {f(x+h)-f(x)}{h}}}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {\frac {f(x+2h)-f(x+h)-f(x+h)+f(x)}{h}}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f(x+2h)-2f(x+h)+f(x)}{h^{2}}}{\Bigg ]}\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1c42fd1e43d33e170ee55c8f860b4446b60df62)
![{\displaystyle {\begin{alignedat}{6}f^{\prime \prime \prime }(x)&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime \prime }(x+h)-f^{\prime \prime }(x)}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime }(x+2h)-2f^{\prime }(x+h)+f^{\prime }(x)}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {{\frac {f(\{x+2h\}+h)-f(x+2h)}{h}}-2{\frac {f(\{x+h\}+h)-f(x+h)}{h}}+{\frac {f(x+h)-f(x)}{h}}}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {\frac {f(x+3h)-f(x+2h)-2f(x+2h)+2f(x+h)+f(x)h)-f(x)}{h}}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}}{\Bigg ]}\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d05cc02953ad7d7483606a523da7845235308a61)
![{\displaystyle \lim _{h\to 0}{\Bigg [}{\frac {f(x+2h)-2f(x+h)+f(x)}{h^{2}}}{\Bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66a916b73926d95ac7ca57b734c7c7cc061bea9b)
![{\displaystyle \lim _{h\to 0}{\Bigg [}{\frac {f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}}{\Bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7118ee80a506ca11adc1834a0f0ac4a6b9893ca5)
![{\displaystyle \lim _{h\to 0}{\Bigg [}{\frac {\sum _{i=0}^{n}{\Big [}(-1)^{i}{\binom {n}{i}}f(x+\{n-i\}h){\Big ]}}{h^{n}}}{\Bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49511aed44e7eb881b7efe0001c451adc1afc4e2)
![{\displaystyle {\begin{aligned}A&=\int _{a}^{b}f(x)\,dx{\mbox{ for individual curves.}}\\&=\int _{a}^{b}{\big [}f(x)-g(x){\big ]}\,dx{\mbox{ for multiple curves.}}\\&=\int _{c}^{d}{\big [}h(y)-k(y){\big ]}\,dy{\mbox{ for multiple curves in respect of the y axis.}}\\V&=\pi \int _{a}^{b}f(x)\,dx{\mbox{ for the Disk Method.}}\\&=\pi \int _{a}^{b}{\big [}f^{2}(x)-g^{2}(x){\big ]}\,dx{\mbox{ for the Washer Method.}}\\&=\int _{a}^{b}A(x)\,dx{\mbox{ with }}A(x)={\mbox{ the area of cross section.}}\\&=2\pi \int _{a}^{b}{\big [}xf^{2}(x){\big ]}\,dx{\mbox{ for Shell Method rotation via the y axis.}}\\&=2\pi \int _{a}^{b}x{\big [}f(x)-g(x){\big ]}\,dx{\mbox{ for Shell Method rotation via the y axis between two curves.}}\\&=\pi \int _{a}^{b}{\big [}({\mbox{outer r}})^{2}-({\mbox{inner r}})^{2}{\big ]}\,dx{\mbox{ for rotation around an alternative axis.}}\\L&=\int _{a}^{b}{\sqrt {1+[f^{\prime }(x)]^{2}}}\,dx{\mbox{ for Curve Length.}}\\&=\int _{a}^{b}{\sqrt {[f^{\prime }(t)]^{2}+[g^{\prime }(t)]^{2}}}\,dt{\mbox{ with Parametric Equations.}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0ee532137605f2be440693890ffb449950d9b65)
![{\displaystyle {\begin{aligned}V(x)&=\int _{-a}^{a}{\bigg [}{\frac {(a^{2}-x^{2})\tan \theta }{2}}{\bigg ]}dx\\&=2\int _{0}^{a}{\bigg [}{\frac {(a^{2}-x^{2})\tan \theta }{2}}{\bigg ]}dx\\&=\int _{0}^{a}{\Big [}(a^{2}-x^{2})\tan \theta {\Big ]}dx\\&=\tan \theta \int _{0}^{a}(a^{2}-x^{2})\,dx\\&=\tan \theta {\Big (}a^{2}x-{\frac {x^{3}}{3}}{\Big )}{\Big |}_{0}^{a}\\&=\tan \theta {\Big (}a^{3}-{\frac {a^{3}}{3}}{\Big )}\\&=a^{3}\tan \theta {\Big (}1-{\frac {1}{3}}{\Big )}\\&=a^{3}\tan \theta {\Big (}{\frac {2}{3}}{\Big )}\\&={\frac {2a^{3}\tan \theta }{3}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5dc886cdf0dc7a7b428bd6c832362d424f1e941)
![{\displaystyle {\begin{aligned}A&=\int _{1}^{\infty }{\frac {1}{x}}\,dx\\&=\lim _{\varphi \to \infty }\int _{1}^{\varphi }{\frac {1}{x}}\,dx\\&=\lim _{\varphi \to \infty }{\big [}\ln {|x|}{\big ]}_{1}^{\varphi }\\&=\lim _{\varphi \to \infty }{\big [}\ln {|\varphi |}-\ln 1{\big ]}\\&=\lim _{\varphi \to \infty }\ln {|\varphi |}\\&=\ln {\infty }\\&=\infty \\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90846da423bed2a43590de1f0a04a54e112d0203)
![{\displaystyle {\begin{aligned}V&=\int _{1}^{\infty }\pi {\Big (}{\frac {1}{x}}{\Big )}dx\\&=\lim _{\aleph \to \infty }\pi \int _{1}^{\aleph }x^{-2}\,dx\\&=\lim _{\aleph \to \infty }{\Big [}-{\frac {\pi }{x}}{\Big ]}_{1}^{\aleph }\\&=\lim _{\aleph \to \infty }{\Big [}-{\frac {\pi }{\aleph }}+{\frac {\pi }{1}}{\Big ]}\\&={\Big [}-{\frac {\pi }{\infty }}+{\frac {\pi }{1}}{\Big ]}\\&=\pi \\&{\mbox{DAMN IT!}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da470b7d8e9c1d6d3564ab40d7023b175969708a)
![{\displaystyle {\begin{array}{lcl}A&=\int _{1}^{\infty }{\frac {1}{x}}\,dxV=&\int _{1}^{\infty }\pi {\Big (}{\frac {1}{x}}{\Big )}dx\\&=\lim _{\varphi \to \infty }\int _{1}^{\varphi }{\frac {1}{x}}\,dx&=\lim _{\aleph \to \infty }\pi \int _{1}^{\aleph }x^{-2}\,dx\\&=\lim _{\varphi \to \infty }{\big [}\ln {|x|}{\big ]}_{1}^{\varphi }&=\lim _{\aleph \to \infty }{\Big [}-{\frac {\pi }{x}}{\Big ]}_{1}^{\aleph }\\&=\lim _{\varphi \to \infty }{\big [}\ln {|\varphi |}-\ln 1{\big ]}&=\lim _{\aleph \to \infty }{\Big [}-{\frac {\pi }{\aleph }}+{\frac {\pi }{1}}{\Big ]}\\&=\lim _{\varphi \to \infty }\ln {|\varphi |}&={\Big [}-{\frac {\pi }{\infty }}+{\frac {\pi }{1}}{\Big ]}\\&=\ln {\infty }&=\pi \\\\&=\infty &{\mbox{DAMN IT!}}\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/add0cd3b0647e4bd27d87a334c5b79cf2d5b4df6)