User:Un Piton

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Dr. Euler's Calculus I Sample Exam Problems

1. True or False

$\int x^{2}\,dx={\frac {1}{3}}x^{3}$

2. True or False

${\mbox{An antiderivative of }}x^{2}{\mbox{ is }}{\frac {1}{3}}x^{3}$

3. True or False

$\int \cos x\,dx=\sin x+c-10$

4. Evaluate

$\int \left\{{\frac {x^{3}-1}{x^{2}}}\right\}\,dx$

${\begin{aligned}&{\mbox{A. }}x^{5}-x^{2}\\&{\mbox{B. }}{\frac {1}{3}}\,(x^{3}-1)^{2}\\&{\mbox{C. }}{\frac {1}{2}}\,x^{2}+{\frac {1}{x}}\\&\color {red}{\mbox{D. }}{\frac {1}{2}}\,x^{2}+{\frac {1}{x}}+c\\&{\mbox{E. }}{\frac {1}{3}}(x^{3}-1)^{2}+c\\\end{aligned}}$

5. Evaluate

$\int {\Big \{}(5x^{2}+1)(5x^{3}+3x-8)^{6}{\Big \}}\,dx$

${\mbox{A. }}{\frac {1}{21}}(5x^{3}+3x-8)^{7}$

${\mbox{B. }}{\frac {1}{21}}(5x^{3}+3x-8)^{7}+c$

${\mbox{C. }}(5x^{2}+1)(5x^{3}+3x-8)^{6}+c$

$\color {red}{\mbox{D. }}{\frac {1}{2}}\,x^{2}+{\frac {1}{x}}+c$

${\mbox{E. }}{\frac {1}{3}}(x^{3}-1)^{2}+c$

6. True or False

$\int x^{n}\,dx={\frac {1}{n+1}}x^{n+1}+c\,{\mbox{ if }}n\neq -1$

7. True or False

$\int {\Big [}f(x){\Big ]}^{n}\,dx={\frac {1}{n+1}}{\Big [}f(x){\Big ]}^{n+1}+c\,{\mbox{ if }}n\neq -1$

8. True or False

$\int {\Big [}f(x){\Big ]}^{n}\,f^{\prime }(x)\,dx={\frac {1}{n+1}}{\Big [}f(x){\Big ]}^{n+1}+c\,{\mbox{ if }}n\neq -1$

9. True or False

$\int (x+7)^{50}\,dx={\frac {1}{51}}(x+7)^{51}+c$

10. True or False

$\int (2x+7)^{50}\,dx={\frac {1}{51}}(2x+7)^{51}+c$

11. True or False

$\int \,dx=x-c$

12. True or False

$\int {\Big [}f(x){\Big ]}^{2}\,dx={\frac {1}{3}}{\Big [}f(x){\Big ]}^{3}+c$

13. Evaluate

$\int {\Big (}x(3x^{2}+1)^{4}{\Big )}\,dx$

${\mbox{A. }}{\frac {1}{5}}(3x^{2}+1)^{5}+c$

${\mbox{B. }}{\frac {x}{5}}(3x^{2}+1)^{5}+c$

$\color {Red}{\mbox{C. }}{\frac {1}{30}}(3x^{2}+1)^{5}+c$

${\mbox{D. }}{\frac {x}{5}}(3x^{2}+1)^{5}$

${\mbox{E. }}{\frac {1}{5}}(3x^{2}+1)^{5}$

14. Evaluate

$\int {\Big (}x^{2}\cos(x^{3}){\Big )}\,dx$

$\color {black}{\mbox{A. }}x^{2}\cos {x^{3}}+c$

${\mbox{B. }}-{\frac {1}{3}}\sin {x^{3}}3x^{3}+c$

${\mbox{C. }}{\frac {1}{3}}\sin {x^{3}}+c$

${\mbox{D. }}{\frac {1}{3}}\sin {x^{3}}3x^{2}+c$

$\color {black}{\mbox{E. }}\cos {3x^{2}}+c$

Evaluation of Multiple Derivatives

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.

$\color {Black}{\mbox{For a monomial to the power of a fraction, or to the power of a negative number}}$

${\mbox{If }}f(x)=ax^{\frac {n_{1}}{n_{2}}}{\mbox{, then }}f^{(m)}(x)={\frac {ax^{{\big (}{\frac {n_{1}}{n_{2}}}-m{\big )}}*{\Bigg [}\prod _{h=1}^{m}{\Big \{}n_{1}-n_{2}(m-h){\Big \}}{\Bigg ]}}{n_{2}^{m}}}{\mbox{ with }}m\in \mathbb {N} {\mbox{ and }}a\neq 0$

$\color {Black}{\mbox{For a monomial to the power of a natural number}}$

${\mbox{If }}f(x)=ax^{n}{\mbox{, then }}f^{(m)}(x)={\Big [}{}_{n}P_{m}{\Big ]}ax^{(n-m)}{\mbox{ with }}n\in \mathbb {N} {\mbox{ and }}m\in \mathbb {N} {\mbox{ and }}a\neq 0$

Dr. Euler's approach follows a different method, using limits as a brute-force method. Let us find multiple derivatives of the function $f(x)$.

$f^{\prime }(x)=\lim _{h\to 0}{\Bigg [}{\frac {f(x+h)-f(x)}{h}}{\Bigg ]}$.
and

${\begin{alignedat}{4}f^{\prime \prime }(x)&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime }(x+h)-f^{\prime }(x)}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {{\frac {f(\{x+h\}+h)-f(x+h)}{h}}-{\frac {f(x+h)-f(x)}{h}}}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {\frac {f(x+2h)-f(x+h)-f(x+h)+f(x)}{h}}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f(x+2h)-2f(x+h)+f(x)}{h^{2}}}{\Bigg ]}\\\end{alignedat}}$
and

${\begin{alignedat}{6}f^{\prime \prime \prime }(x)&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime \prime }(x+h)-f^{\prime \prime }(x)}{h}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f^{\prime }(x+2h)-2f^{\prime }(x+h)+f^{\prime }(x)}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {{\frac {f(\{x+2h\}+h)-f(x+2h)}{h}}-2{\frac {f(\{x+h\}+h)-f(x+h)}{h}}+{\frac {f(x+h)-f(x)}{h}}}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {\frac {f(x+3h)-f(x+2h)-2f(x+2h)+2f(x+h)+f(x)h)-f(x)}{h}}{h^{2}}}{\Bigg ]}\\&=\lim _{h\to 0}{\Bigg [}{\frac {f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}}{\Bigg ]}\\\end{alignedat}}$

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.

Suppose you have a function f(x). The derivatives of of this function are as follow:

Derivative	Prime Notation	Limit notation	Binomial coefficients
0	$\color {Black}f(x)$	$\lim _{h\to 0}f(x)$	1
$1^{st}$	$f^{\prime }(x)$	$\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$	1 1
$2^{nd}$	$f^{\prime \prime }(x)$	$\lim _{h\to 0}{\Bigg [}{\frac {f(x+2h)-2f(x+h)+f(x)}{h^{2}}}{\Bigg ]}$	1 2 1
$3^{rd}$	$f^{\prime \prime \prime }(x)$	$\lim _{h\to 0}{\Bigg [}{\frac {f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}}{\Bigg ]}$	1 3 3 1
$\vdots$	$\vdots$	$\vdots$	$\vdots$
$n^{th}$	$\color {black}f^{(n)}(x)$	$\lim _{h\to 0}{\Bigg [}{\frac {\sum _{i=0}^{n}{\Big [}(-1)^{i}{\binom {n}{i}}f(x+\{n-i\}h){\Big ]}}{h^{n}}}{\Bigg ]}$	1 n ${\binom {n}{2}}\,\dots \,{\binom {n}{n-2}}$ n 1

Calculus II Equations

${\begin{aligned}A&=\int _{a}^{b}f(x)\,dx{\mbox{ for individual curves.}}\\&=\int _{a}^{b}{\big [}f(x)-g(x){\big ]}\,dx{\mbox{ for multiple curves.}}\\&=\int _{c}^{d}{\big [}h(y)-k(y){\big ]}\,dy{\mbox{ for multiple curves in respect of the y axis.}}\\V&=\pi \int _{a}^{b}f(x)\,dx{\mbox{ for the Disk Method.}}\\&=\pi \int _{a}^{b}{\big [}f^{2}(x)-g^{2}(x){\big ]}\,dx{\mbox{ for the Washer Method.}}\\&=\int _{a}^{b}A(x)\,dx{\mbox{ with }}A(x)={\mbox{ the area of cross section.}}\\&=2\pi \int _{a}^{b}{\big [}xf^{2}(x){\big ]}\,dx{\mbox{ for Shell Method rotation via the y axis.}}\\&=2\pi \int _{a}^{b}x{\big [}f(x)-g(x){\big ]}\,dx{\mbox{ for Shell Method rotation via the y axis between two curves.}}\\&=\pi \int _{a}^{b}{\big [}({\mbox{outer r}})^{2}-({\mbox{inner r}})^{2}{\big ]}\,dx{\mbox{ for rotation around an alternative axis.}}\\L&=\int _{a}^{b}{\sqrt {1+[f^{\prime }(x)]^{2}}}\,dx{\mbox{ for Curve Length.}}\\&=\int _{a}^{b}{\sqrt {[f^{\prime }(t)]^{2}+[g^{\prime }(t)]^{2}}}\,dt{\mbox{ with Parametric Equations.}}\\\end{aligned}}$

The Volume of a Cylindrical Wedge

For a cylindrical wedge, similar to the type of wedge that is cut from a tree when chopping, this mathematical equation will find the volume. If you take a cross section of the wedge to form a right triangle, the area of that cross section can be computed as follows, with $a$ representing the radius of of the semicircle, $x$ representing the position on the graph of the ellipse, and $\theta$ representing the angle the wedge is cut.

${\begin{aligned}A(x)&={\frac {{\Big (}{\sqrt {a^{2}-x^{2}}}{\Big )}{\Big (}{\sqrt {a^{2}-x^{2}}}\tan \theta {\Big )}}{2}}\\&={\frac {(a^{2}-x^{2})\tan \theta }{2}}\\\end{aligned}}$
From here, we can calculate the volume of the wedge using the Cross Section method in calculus.

${\begin{aligned}V(x)&=\int _{-a}^{a}{\bigg [}{\frac {(a^{2}-x^{2})\tan \theta }{2}}{\bigg ]}dx\\&=2\int _{0}^{a}{\bigg [}{\frac {(a^{2}-x^{2})\tan \theta }{2}}{\bigg ]}dx\\&=\int _{0}^{a}{\Big [}(a^{2}-x^{2})\tan \theta {\Big ]}dx\\&=\tan \theta \int _{0}^{a}(a^{2}-x^{2})\,dx\\&=\tan \theta {\Big (}a^{2}x-{\frac {x^{3}}{3}}{\Big )}{\Big |}_{0}^{a}\\&=\tan \theta {\Big (}a^{3}-{\frac {a^{3}}{3}}{\Big )}\\&=a^{3}\tan \theta {\Big (}1-{\frac {1}{3}}{\Big )}\\&=a^{3}\tan \theta {\Big (}{\frac {2}{3}}{\Big )}\\&={\frac {2a^{3}\tan \theta }{3}}\\\end{aligned}}$

The Paradox of Gabriel's Horn

${\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}}$${\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}}$ ${\begin{array}{lcl}A&=\int _{1}^{\infty }{\frac {1}{x}}\,dxV=&\int _{1}^{\infty }\pi {\Big (}{\frac {1}{x}}{\Big )}dx\\&=\lim _{\varphi \to \infty }\int _{1}^{\varphi }{\frac {1}{x}}\,dx&=\lim _{\aleph \to \infty }\pi \int _{1}^{\aleph }x^{-2}\,dx\\&=\lim _{\varphi \to \infty }{\big [}\ln {|x|}{\big ]}_{1}^{\varphi }&=\lim _{\aleph \to \infty }{\Big [}-{\frac {\pi }{x}}{\Big ]}_{1}^{\aleph }\\&=\lim _{\varphi \to \infty }{\big [}\ln {|\varphi |}-\ln 1{\big ]}&=\lim _{\aleph \to \infty }{\Big [}-{\frac {\pi }{\aleph }}+{\frac {\pi }{1}}{\Big ]}\\&=\lim _{\varphi \to \infty }\ln {|\varphi |}&={\Big [}-{\frac {\pi }{\infty }}+{\frac {\pi }{1}}{\Big ]}\\&=\ln {\infty }&=\pi \\\\&=\infty &{\mbox{DAMN IT!}}\\\end{array}}$