Q. 15 - 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→∞.

${\int }_{-2}^3\left(4-x^2\right) dx$

Step-by-Step Solution

Answer

The definite integral is $\frac{25}{3}$ .

Step 1. Given information .

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

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 integral.

$$\begin{gathered} \delta x=\frac{3+2}{n}=\frac{5}{n} \\ a+k·\delta x=-2+k\frac{5}{n} \\ f\left(a+k·\delta x\right)·\delta x=\left[4-{\left(\frac{5k}{n}-2\right)}^2\right]·\frac{5}{n} \\ =\frac{100k}{n^2}-\frac{125k^2}{n^3} \\ {\int }_a^{b}f\left(x\right) dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(a+k·\delta x\right)·\delta x \\ =\underset{n\to \infty }{\lim }\frac{100}{n^2}\sum _{k=1}^{n}k-\frac{125}{n^3}\sum _{k=1}^n{k}^2 \\ =\underset{n\to \infty }{\lim }\frac{100}{n^2}·\frac{n\left(n+1\right)}{2}-\frac{125}{n^3}·\frac{n\left(n+1\right)\left(2n+1\right)}{6} \\ =\underset{n\to \infty }{\lim }\frac{100}{n^2}·\frac{n^2\left(1+{\displaystyle \frac{1}{n}}\right)}{2}-\frac{125}{n^3}·\frac{n^3\left(1+{\displaystyle \frac{1}{n}}\right)\left(2+{\displaystyle \frac{1}{n}}\right)}{6} \\ =\frac{25}{3} \end{gathered}$$