How do I apply the Riemann sum with equal-width sub-intervals?

Question

$$\int^b_af(x)dx=\lim_{n\rightarrow\infty} \sum^n_{k=1}f\left(a+k\cdot\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right)$$

I am trying to apply this to $$\int_0^4 (4x+2) dx$$

In this scenario, what is $n$? is $n=4$?

Answer 1

$n$ is not $4$. $n$ is not fixed. You see the limit definition: that means you must consider the quantity, the Riemann (partial) sum, for every $n$ and (if this sequence has a limit) find its limit.

So, for some $n\ge1$, we calculate: $$\sum_{k=1}^nf\left(a+k\frac{b-a}{n}\right)\frac{b-a}{n}=\sum_{k=1}^n\left(4\left(0+k\frac{4-0}{n}\right)+2\right)\frac{4-0}{n}$$You should go from here. Calculate this sum exactly (this is possible) and then let $n\to\infty$.

You should use: $$\sum_{k=1}^n k=\frac{n(n+1)}{2}$$

Answer 2

Just use the definition what you posted.

• The interval is given by $[a,b]=[0,4]$, that is, $a:=0$ and $b:=4$.
• If the function is $f(x)=4x+2$, then $$f\left(\underbrace{a+k\frac{b-a}{n}}_{=x}\right)=4\left(0+k\frac{4-0}{n}\right)+2=4^{2}\left(\frac{k}{n}\right)+2.$$
• Also, $$\left(\frac{b-a}{n}\right)=\frac{4-0}{n}=\frac{4}{n}.$$
• Hence, \begin{align*} \int_{0}^{4}(4x+2)&=\lim_{n\to+\infty}\sum_{k=1}^{n}\left(\frac{4^{2}k}{n}+2\right)\left(\frac{4}{n}\right),\\&=\lim_{n\to+\infty}\sum_{k=1}^{n}\left(\frac{4(2n+4^{2}k)}{n^2}\right),\\ &=\lim_{n\to+\infty}\sum_{k=1}^{n}\left(\frac{8}{n}+\frac{4^3k}{n^{2}}\right), \\&=\lim_{n\to+\infty}\left(\sum_{k=1}^{n}\underbrace{\frac{8}{n}}_{\text{constant respect to $k$}}+\sum_{k=1}^{n}\underbrace{\frac{4^{3}}{n^{2}}}_{\text{constant respect to $k$}}k\right), \\ &=\lim_{n\to+\infty}\left(\frac{8}{n}\sum_{k=1}^{n}1+\frac{4^{3}}{n^{2}}\sum_{k=1}^{n}k\right),\\ &=\lim_{n\to+\infty}\left(\frac{8}{n}\cdot n+\frac{4^{3}}{n^{2}}\cdot \frac{n(n+1)}{2}\right),\\&=40. \end{align*}