Q. 14 - Calculus

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→∞.

${\int }_0^4{x}^2 dx$

Step-by-Step Solution

Answer

The definite value is $\frac{64}{3}$.

Step 1. Given information .

Consider the given definite integral ${\int }_0^4{x}^2 dx$ .

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. Calculating the integral .

$$\begin{gathered} \delta x=\frac{4-0}{n}=\frac{4}{n} \\ f\left(a+k·\delta x\right)={\left(\frac{4k}{n}\right)}^2·\frac{4}{n}=\frac{64k^2}{n^3} \\ {\int }_a^{b}f\left(x\right) dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(a+k·\delta x\right)\delta x \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{64}{n^3}·k^2 \\ =\underset{n\to \infty }{\lim }\frac{64}{n^3}·\frac{n\left(n+1\right)\left(2n+1\right)}{6} \\ =\underset{n\to \infty }{\lim }\frac{64}{n^3}·\frac{n^3\left(1+{\displaystyle \frac{1}{n}}\right)\left(2+{\displaystyle \frac{1}{n}}\right)}{6} \\ =\frac{64}{3} \end{gathered}$$