I need to prove that lim (mx+b) as x approaches c = mc +b. Any tips or help would be greatly appreciated.
Also, What exactly is the relationship of the definite integral to the derivative?
Again, help and or tips would be greatly appreciated.
I need to prove that lim (mx+b) as x approaches c = mc +b. Any tips or help would be greatly appreciated.
Also, What exactly is the relationship of the definite integral to the derivative?
Again, help and or tips would be greatly appreciated.
Hello,
For the second one, if f is a function and f ' is its derivative, we have :
$\displaystyle \int_a^b f'(x) ~dx=f(b)-f(a)$
Usually, we define an antiderivative of f, F and say $\displaystyle \int_a^b f(x) ~dx=F(b)-F(a)$
Since f is an antiderivative of f ', we have the formula above.
This actually is not that bad. I know that the epsilon-delta thing can be confusing.
$\displaystyle f(x)=mx+b, \;\ m\neq 0$
$\displaystyle \lim_{x\to c}(mx+b)=mc+b$
$\displaystyle |(mx+b)-(mc+b)|<{\epsilon}$
$\displaystyle |m(x-c)|<{\epsilon}$
$\displaystyle |x-c|<\frac{\epsilon}{|m|}={\delta}$
Therefore, for $\displaystyle {\epsilon}>0$, let $\displaystyle {\delta}=\frac{\epsilon}{|m|}$
An alternative to Moo's post would be that in a sense integration and differentiation are inverses. If $\displaystyle F(x)=\int_{a}^x f~dx$ and $\displaystyle f$ is continuous at some point $\displaystyle x_0$ then $\displaystyle F(x)$ is differentiable at $\displaystyle x_0$ and $\displaystyle F'(x_0)=f(x_0)$