Question

Calculating definite integrals with limits of Riemann sums: Calculate

the exact value of each the following definite integrals by

setting up a general Riemann sum and then taking the limit

as n→∞.

${\int }_1^4{\left(3x+1\right)}^{2}dx$

Short answer

The definite integral is 237 .

Step-by-step solution

Step 1. Given information .

Consider the given integral${\int }_1^4{\left(3x+1\right)}^{2}dx$.

Step 2. Formula used .

${\int }_a^{b}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(a+k·\delta x\right)·\delta x$

Step 3. Find the definite integral .

$$\begin{gathered} \delta x=\frac{4-1}{n}=\frac{3}{n} \\ a+k·\delta x=1+k·\frac{3}{n} \\ f\left(a+k·\delta x\right)={\left[3\left(1+\frac{3k}{n}\right)+1\right]}^2=16+\frac{81k^2}{n^2}+\frac{72k}{n} \\ f\left(a+k·\delta x\right)·\delta x=\frac{48}{n}+\frac{243}{n^3}·k^2+\frac{216k}{n^2} \\ {\int }_a^{b}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(a+k·\delta x\right)·\delta x \\ =\underset{n\to \infty }{\lim }\frac{48}{n}\sum _{k=1}^{n}1+\frac{243}{n^3}\sum _{k=1}^n{k}^2+\frac{216}{n^2}\sum _{k=1}^{n}k \\ =\underset{n\to \infty }{\lim }\frac{48}{n}·n+\frac{243}{n^3}·\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{216}{n^2}·\frac{1}{2} \\ =237 \end{gathered}$$