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Mean Value Theorem

Tags

Calculus

Cegep/1

Cegep/2

Word count

212 words

Reading time

2 minutes

Abbr. M. V. T.

Let $f(x)$ be a function continuous over $[a,b]$ and differentiable over $(a,b)$,
then $\mathrm{\exists }c\in [a,b]$ such that

$$\begin{array}{rl}{f}^{\prime }(c)& =\frac{f(b)-f(a)}{b-a}\\ f(c)& =\frac{1}{b-a}{\int }_a^{b}f(x)\mathrm{d}x\end{array}$$

Proof for integrals

We know that $f$ is continuous on $[a,b]$.
Define $A(x)={\int }_a^{x}f(t)\mathrm{d}t$, then by FTC1 we get:

1. $A(x)$ is continuous on $[a,b]$.
2. $A(x)$ is differentiable on $(a,b)$.
3. $A^{\prime }(x)=f(x)$

By 1. and 2. we can apply MVT to $A(x)$ on $[a,b]$ to get that $\mathrm{\exists }c\in [a,b]$ such that

$$A^{\prime }(c)=\frac{A(b)-A(a)}{b-a}$$

Using 3.,

$$f(c)=$$

Note that $A(b)={\int }_a^{b}f(t)$ and $A(a)={\int }_a^{a}f(t)\mathrm{d}t=0$.

$$=\frac{1}{b-a}{\int }_a^{b}f(t)\mathrm{d}t$$

$◻$

Examples

If a steel ball is dropped, its velocity at time $t$ is approximated by $v(t)=-10t\frac{\mathrm{m}}{\mathrm{s}}$. Find the average velocity between the time it is dropped at 373m and when it hits the ground.

We find $t_2$ by solving $x(t)=0$.

$$\begin{array}{rl}x(t)& =\int v(t)\mathrm{d}t\\ & =-5t^2+373\end{array}$$

Contributors

Ajitani Hifumi

Changelog

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