Write expnential functions in the form of y =ae^(kt) find k accurate to four decimal places. If t is measured inyears, give the percent annual growth rate and continuous growthrate per year. a) a city is growing by 26% per year. b)a company's profit is increasing by an annual growthfactor of 1.12

Existence of antiderivative without Cauchy-Goursat's theorem
I wonder if anybody has tried the following kind of direct proof for the existence of an antiderivative of an analytic function on a star-shaped domain.
Theorem: Let $f:D\to \mathbb{C}$ be an analytic function on a star-shaped domain D. Then f has an antiderivative F on D.
"proof": For simplicity, assume that every point in D is connected to $0\in \mathbb{C}$ by a line segment. Define
$F(z)={\int }_0^{1}zf(zt)dt.$
Then $\underset{h\to 0}{lim}\left(\frac{F(z+h)-F(z)}{h}\right)=\underset{h\to 0}{lim}{\int }_0^1\left(\frac{(z+h)f(zt+ht)-zf(t)}{h}\right)dt.$
It can be checked that as $h\to 0$, the integrand converges to $\frac{d}{dt}tf(zt).$.
Therefore, $F^{\prime }(z)=f(z)$, provided that the limit and integral are interchangeable. qed.
Of course, a limit and integral cannot always be interchanged. But I wonder if anybody seriously considered the above line of proof.
Thanks. As far as I can search from several textbooks in complex variables, the above theorem is proved by using Cauchy-Goursat's theorem. More concretely, they use the equality ${\int }_{z_0}^{z+h}f(z)dz-{\int }_{z_0}^{z}f(z)dz={\int }_z^{z+h}f(z)dz,$, which can be justified by Cauchy-Goursat's theorem. The point of my question is: Is it possible to directly invoke to the computation as above, without using Cauchy-Goursat's theorem? If this is possible, we get another proof of Cauchy-Goursat's theorem, at least for star-convex domains.

What is the equation of the line normal to ${\displaystyle f\left(x\right)=x^2-x-2}$ at x=2?

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Simplifying Integral Expression w/ Antiderivative
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How to following limit
$\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{n^2}+\frac{2}{n^2}+\cdots +\frac{(n-1)}{n^2})(\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{n+n}).$