Assignment-daixieTM为您提供斯坦福大学Stanford University MATH 20 Calculus微积分代写代考和辅导服务！
Instructions:
This is a course in integral calculus, which is the second part of calculus after differential calculus. It covers the following topics:
- Definite Integral: Students will learn how to compute the definite integral of a function over a given interval using Riemann sums, which is a technique for approximating the area under a curve.
- Antiderivatives and Fundamental Theorem of Calculus: Students will learn how to find the antiderivative of a function, which is the inverse operation of differentiation. They will also learn the Fundamental Theorem of Calculus, which establishes the relationship between differentiation and integration.
- Integration Techniques: Students will learn integration by substitution and integration by parts, which are techniques for simplifying the integration process.
- Applications of Integration: Students will learn how to use integration to find the area between two curves, and the volume of solids of revolution using slices, washers, and shells.
- Differential Equations: Students will learn how to solve initial-value problems, which involve finding the solution to a differential equation given an initial condition. They will also learn about exponential and logistic models, direction fields, and parametric curves.
Prerequisite for this course is Math 19 or equivalent. If the student has not previously taken a calculus course at Stanford, then they must have taken the math placement diagnostic offered through the Math Department website in order to register for this course.
Let $M>0$ and $f(x)=x^3$ for $0 \leq x \leq M$. Find a value of $c$ which satisfies the conclusion of the mean value theorem for the function $f$ over the interval $[0, M]$. Answer: $c=\frac{M}{a}$___ where $a=$__
By the Mean Value Theorem, there exists $c \in (0,M)$ such that $$f'(c)=\frac{f(M)-f(0)}{M-0}=\frac{M^3-0^3}{M} = M^2.$$ Since $f(x)=x^3$, we have $f'(x)=3x^2$. So, we have $f'(c)=3c^2=M^2$. Solving for $c$ gives $c=\frac{M}{\sqrt{3}}$.
Therefore, $a=\sqrt{3}$ and $c=\frac{M}{\sqrt{3}}$.
Let $f(x)=x^4+x+3$ for $0 \leq x \leq 2$. Find a point $c$ whose existence is guaranteed by the mean value theorem. Answer: $c=2^p$ ____where $p=$___
By the Mean Value Theorem, there exists $c \in (0,2)$ such that $$f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{(2^4+2+3)-(0^4+0+3)}{2}=17.$$ Since $f(x)=x^4+x+3$, we have $f'(x)=4x^3+1$. So, we have $f'(c)=4c^3+1=17$. Solving for $c$ gives $c= \sqrt[3]{4}$.
Therefore, $p=\frac{\ln(4)}{\ln(2)}$. Hence, $c=2^p=\boxed{2^{\frac{\ln(4)}{\ln(2)}}}$.
Let $f(x)=\sqrt{x}$ for $4 \leq x \leq 16$. Find a point $c$ whose existence is guaranteed by the mean value theorem. Answer: $c=$______
By the Mean Value Theorem, there exists $c \in (4,16)$ such that $$f'(c)=\frac{f(16)-f(4)}{16-4}=\frac{\sqrt{16}-\sqrt{4}}{12}=\frac{1}{3}.$$ Since $f(x)=\sqrt{x}$, we have $f'(x)=\frac{1}{2\sqrt{x}}$. So, we have $f'(c)=\frac{1}{2\sqrt{c}}=\frac{1}{3}$. Solving for $c$ gives $c= \frac{9}{4}$.
Therefore, $c=\boxed{\frac{9}{4}}$.
