Trouble evaluating definite integral with a Riemann sum $$ \int_{1}^{9}(x^2+7) dx $$
That's the integral. And I'm supposed to evaluate it by using:
$$ \lim_ {x \to \infty} \sum_{i=1}^{n} f(x_i)(\Delta x) $$
Also $$ \Delta x = \frac{b-a}{n} $$
And $$ x_i = a + \frac{i(b-a)}{n} $$
The question asks: "Now find the sum of the areas of
n approximating rectangles. (Your answer will include variable n)"
My solution:
$$ \lim_ {n \to \infty} \sum_{i=1}^{n} \left[ \left(\frac{8i}{n}\right)^2 + 7\left(\frac{8i}{n}\right) \right] \cdot \frac{8}{n} $$
$$\lim_ {n \to \infty} \left(\left(\frac{8}{n}\right)^2 \cdot \frac{8}{n}\right) \sum_{i=1}^{n} [i^2] + \left(\left(\frac{8}{n}\right) \cdot \frac{8}{n}\right)\sum_{i=1}^{n} [ i ] $$
$$ \lim_ {n \to \infty} \left(\frac{8}{n}\right)^3 \left(\frac{n(n+1)}{2}\right)^2 + \left(\frac{8}{n}\right)^2\left(\frac{n(n+1)}{2}\right) $$
I've tried a few variations of this but nothing seems to work... The final answer is supposed to be 896/3 and the question I cited earlier calls for a simplified version of what I have up there, but this just isn't making sense at all.
 A: The formula of the sum of first $n$ squares is mistakenly replaced by that of the sum of first $n$ cubes.

Using the elementary definition of the definite integral,
$$
\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{b-a}{n} f\left(a+\frac{k(b-a)}{n}\right)
$$
Putting $a=1$ and $b=9$ and $f(x)=x^2+7$ yields
$$
\begin{aligned}
\int_1^9\left(x^2+7\right) d x & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{8}{n}\left[\left(1+\frac{8 k}{n}\right)^2+7\right] \\
& =64 \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\frac{1}{n}+\frac{2 k}{n^2}+\frac{8 k^2}{n^3}\right)
\end{aligned}
$$
Using the formula for sum of first $n$ positive integers and their squares,
$$
\sum_{k=1}^n k=\frac{n}{2}(n+1) \text { and } \sum_{k=1}^n k^2=\frac{n}{6}(n+1)(2 n+1),
$$
we get $$
\begin{aligned}
\int_1^9\left(x^2+7\right) d x & =64 \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\frac{1}{n}+\frac{2 k}{n^2}+\frac{8 k^2}{n^3}\right) \\
& =64 \lim _{n \rightarrow \infty}\left(1+\frac{2}{n^2} \cdot \frac{n(n+1)}{2}+\frac{8}{n^3}\cdot \frac{n}{6}(n+1)(2n+1)\right) \\
& =64\left(2+\frac{8}{3}\right) \\
& =\frac{896}{3}
\end{aligned}
$$
A: You used cubic sum.??
I usually use equal-partition left Riemann sum defined by
$$\int_a^bf(x)dx=\lim_{n\rightarrow \infty}\sum_{i=0}^{n-1}f(x_i)\Delta x$$
where $x_i=a+i\Delta x$ and $\Delta x=\frac{b-a}{n}$. For this question, $\Delta x=\frac{9-1}{n}=\frac{8}{n}$ and $x_i=\frac{8i}{n}+1.$ Hence,
$$\int_1^9(x^2+7)dx=\lim_{n\rightarrow \infty}\sum_{i=0}^{n-1}\left(\left(\frac{8i}{n}+1\right)^2+7\right)\left(\frac{8}{n}\right)$$
You made something wrong here also.
$$\begin{align}\int_1^9(x^2+7)dx&=\lim_{n\rightarrow \infty}\sum_{i=0}^{n-1}\left(\frac{64}{n^2}i^2+\frac{16}{n}i+8\right)\left(\frac{8}{n}\right)\\
&=\lim_{n\rightarrow \infty}\frac{512}{n^3}\sum_{i=0}^{n-1}i^2+\frac{128}{n^2}\sum_{i=0}^{n-1}i+\frac{64}{n}\sum_{i=0}^{n-1}1\\
&=\lim_{n\rightarrow \infty}\frac{512}{n^3}\frac{(n-1)n(2n-1)}{6}+\frac{128}{n^2}\frac{(n-1)n}{2}+\frac{64}{n}n\\
&=\frac{512}{3}+64+64\\
&=\frac{896}{3}
\end{align}$$
