Estimating the Sum of Square Roots (Calculus Starter)? I am trying to estimate the value of the Sum below without using the calculator:
$$
\sum_{n=1}^{10000} \sqrt{i}
$$
I have looked for other ways around and tried to turn it into  $n\sqrt{n}\int_{0}^{1} \sqrt{x} dx$, but It doesn't work quite well ( the estimation isn't really close to the actual answer computed by the calculator ). Is there a comprehensible method for a calculus starter like me?
Thanks!
 A: Since $x\to \sqrt{x}$ is strictly increasing, then for $n\geq 1$,
$$\sum_{i=1}^{n-1} \sqrt{i}<\underbrace{\int_0^{n}\sqrt{x}\,dx}_{=\frac{2}{3}n\sqrt{n}}< \sum_{i=1}^{n} \sqrt{i}.$$
Hence, by letting $n=10000$, we find that $\sqrt{n}=100$ and
$$666666.666<\frac{2000000}{3}<\sum_{i=1}^{10000} \sqrt{i}< \frac{2000000}{3}+100<666766.667.$$
Since $x\to \sqrt{x}$ is also strictly concave, we can find a much better upper bound by using trapezoids instead of rectangles:
$$\sum_{i=0}^{n-1} \frac{\sqrt{i}+\sqrt{i+1}}{2}< \int_0^{n}\sqrt{x}\,dx=\frac{2}{3}n\sqrt{n}$$
which leads to
$$\sum_{i=1}^{10000} \sqrt{i}<\frac{2000000}{3}+50<666716.667.$$
Note that $\sum_{i=1}^{10000} \sqrt{i}\approx 666716.459$, so the above approximation is not too bad (by the above bounds we know that the absolute error is less than $50$).
A: if we have $f(x) > 0$ and $f'(x) > 0,$  then
$$ \int_{a-1}^{b} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a}^{b+1} \;  f(x) \; dx   $$
These are about bounding $f(n)$   above and below by an integral.  We do better by integrating the function from $n - \frac{1}{2} $ to
$n + \frac{1}{2} $
Either way, we do better with
$$  \; \sum_{j=a}^b \;  g(j) \; \approx \; \int_{a-\frac{1}{2}}^{b+\frac{1}{2} } \;  g(x) \; dx   $$
As the size of the error depends on the second derivative of $f,$   this comes out better. The second derivative gets small as the argument gets large. A good deal of error occurs ate the beginning, so we may take $a=10,$ for example, and carefully add on the square roots of the numbers from $1$ to $9$
First,
$$  \; \sum_{j=1}^{10000} \;  \sqrt j \; \approx \; \int_{1-\frac{1}{2}}^{10000+\frac{1}{2} } \;  \sqrt x \; dx   $$
Second try,
$$  \; \sum_{j=10}^{10000} \;  \sqrt j \; \approx \; \int_{10- \frac{1}{2}}^{10000+\frac{1}{2} } \;  \sqrt x \; dx   $$
so
$$  \; \sum_{j=1}^{10000} \;  \sqrt j \; \approx \;  \; \sum_{j=1}^{9} \;  \sqrt j \; \; + \; \; \int_{10- \frac{1}{2}}^{10000+\frac{1}{2} } \;  \sqrt x \; dx   $$
