Q. 13 - 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→∞.

${\int }_3^5\left(2x+4\right) dx$

Step-by-Step Solution

Answer

The exact value is 24 .

Step 1. Given information .

Consider the given integral ${\int }_3^5\left(2x+4\right) dx$ .

Step 2. Formula used .

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

Step 3. Find the value .

$$\begin{gathered} {\int }_3^5\left(2x+4\right) dx=\delta x=\frac{5-3}{n}=\frac{2}{n} \\ a+k·\delta x=3+k\frac{2}{n} \\ f\left(a+k·\delta x\right) ·\delta x=\left[2\left(3+\frac{2k}{n}\right)\right]·\frac{2}{n} \\ =\left(\frac{4k}{n}+1\right)·\frac{2}{n} \\ {\int }_a^{b}f\left(x\right) dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(a+k·\delta x\right)·\delta x \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(\frac{8k}{n^2}+\frac{20}{n}\right) \\ =\frac{8}{n^2}·\frac{n\left(n+1\right)}{2}+\frac{20}{n}·n \\ =24 \end{gathered}$$