0
$\begingroup$

enter image description here

I am struggling to understand the last parts of this proof because I know that the IVT states that on the interval $[a,b]$ of $f$, where it is continuous, there exists a value $L$ between $f(a)$ and $f(b)$ such that $f(c)=L$. Now, while I understand that, I do not singularly understand how a single value of a function, $f(k)$, can be equal to the the area of our function multiplied by $\frac{1}{b-a}$, for, as I know, the IVT can be applied only to points of a function and not integrals. Can anyone tell where I have gone wrong in my understanding?

$\endgroup$
1
$\begingroup$

The IVT does not say "there exists a value $L$ between $f(a)$ and $f(b)$ such that $f(c)=L$". It says that if $L$ is any value between $f(a)$ and $f(b)$ then there exists $c\in(a,b)$ such that $f(c)=L$. Note the "any".

Consider the interval $[m,M]$. Let $L=\frac{1}{b-a}\int_a^b f(t)\,dt$. Then $L$ is a value between $f(m)$ and $f(M)$ (one of the inequalities above says so). So IVT says exactly that there exists $k\in(m,M)$ with $f(k)=L$. (And now $k\in(a,b)$ since $(m,M)\subset(a,b)$.

(I assumed here that $m<M$ just so I could use the notation (m,M)$. If $M

$\endgroup$
0
$\begingroup$

The function in question here is the anti-derivative of $f(x)$ starting at $a$.
$$F(x) = \int_a^x f(x) dx$$ This function is zero at $F(a)$ and equal to $\int_a^b f(x) dx$ at $F(b)$. At an intermediate value between $a$ and $b$ it will have differing values, but there is a point in between ($k$) where everything balances out: $$\int_a^b f(x) dx = f(k)(b-a)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.