Calculate $\int^3_{0}\frac{-x^2}{4}+5dx$ by using the definition $\int^b_{a}f(x)dx=\lim_{n \to \infty}[ \overset{n} \sum_{i = 1} f(x_{i})\Delta x]$?

Question

I am trying to calculate $\int^3_{0}\frac{-x^2}{4}+5dx$ by using the definition $\int^b_{a}f(x)dx=\lim_{n \to \infty}[ \overset{n} \sum_{i = 1} f(x_{i})\Delta x]$.

The full question text is:

consider the function $f(x)= - \frac{x^2}{4}+5$ by using the definition

$$\int^b_{a}f(x)dx=\lim_{n \to \infty}[ \overset{n} \sum_{i = 1} f(x_{i})\Delta x]$$

The summation inside the brackets is $R_{n}$ which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

Calculate $R_{n}$ for $f(x)= -\frac{x^2}{4}+5$ on the interval $[0, 3]$ and write your answer as a function of $n$ without any summation signs. You will need the summation formulas

$$\overset{n}\sum_{i = 1} i = \frac{n(n+1)}{2}$$

$$\overset{n}\sum_{i = 1} i^2 = \frac{n(n+1)(2n+1)}{6}$$

$R_n=$ ?

$\lim_{n \to \infty} R_n=$ ?

I am not sure why they are giving me the summation formulas here.

I thought problems like this were solved by finding $\Delta x$ and then using that to find $x_{i}$.

Then I would plug in $x_{i}$ to the formula $\lim_{n \to \infty}[ \overset{n} \sum_{i = 1} f(x_{i})\Delta x]$, and simplify.

It seems like to use the summation formulas mentioned, I would need to get everything but maybe $x_{i}$ to the left the $\sum$ symbol, but I'm not sure how, or if that's what I'm supposed to do.

Since $\Delta x = \frac{b-a}{n}=\frac{3}{n}$ and $x_i = a - \Delta x_i = \frac{3_i}{n}$

I thought the answer would come from $\lim_{n \to \infty}[ \overset{n} \sum_{i = 1} (\frac{3_i}{n})\frac{3}{n}]$, but I'm not sure what else I can do with this, and I don't know what form of answer the question is asking for.

Answer 1

Using the definition given you have $$f(x_i)=f\left(\frac{3i}n\right)=-\frac{9i^2}{4n^2}+5$$ $$\Delta x=\frac3n$$ Hence the integral is given by $$\begin{align} \lim_{n\to\infty}\sum_{i=1}^n\left(\left(-\frac{9i^2}{4n^2}+5\right)\frac3n\right) &=\lim_{n\to\infty}\sum_{i=1}^n\left(-\frac{27i^2}{4n^3}+\frac{15}n\right)\\ &=\lim_{n\to\infty}\left(-\frac{27}{4n^3}\sum_{i=1}^ni^2+\frac{15}n\sum_{i=1}^n 1\right)\\ &=\lim_{n\to\infty}\left(-\frac{27}{4n^3}\left(\frac{n(n+1)(2n+1)}6\right)+\frac{15}n(n)\right)\\ &=\lim_{n\to\infty}\left(-\frac{27}{4}\left(\frac{(1+\frac1n)(2+\frac1n)}6\right)+15\right)\\ &=-\frac{27}4\left(\frac26\right)+15\\ &=-\frac94+15\\ &=\frac{51}4\\ \end{align}$$

Answer 2

Hint:

You seem to have all the elements. By linearity, you have to determine \begin{cases}\displaystyle \sum _{i=1}^n x_i^2\,\Delta x = \sum _{i=1}^n \Bigl(\frac{3i}{n}\Bigr)^2\frac 3n=\frac{27}{n^3}\sum _{i=1}^n i^2,\\[1ex] \displaystyle\sum _{i=1}^n 1\,\Delta x=3. \end{cases} Can you continue?