Q. 14 - Calculus

Question

Calculate each definite integral in Exercises 13–14, using (a) the definition of the definite integral as a limit of Riemann sums, (b) the definite integral formulas from Theorem 4.13, and (c) the Fundamental Theorem of Calculus. Then show that your three answers are the same 

${\int }_0^4{\left(3x+2\right)}^{2}dx$

Step-by-Step Solution

Answer

Part (a) ${\int }_0^4{\left(3x+2\right)}^{2}dx=76$

Part (b)  ${\int }_0^4{(3x+2)}^{2}dx=304$

Part (c)   ${\int }_0^4{(3x+2)}^{2}dx=304$

Part (a) Step 1. Given information

The given integral ${\int }_0^4{\left(3x+2\right)}^{2}dx$

Part (a) Step 2. The definition of the definite integral as a limit of Riemann sums

Let's take 

$$\begin{gathered} ∆x=\frac{b-a}{n} \\ =\frac{4-0}{n} \\ =\frac{4}{n} and x_i=\frac{4i}{n} \end{gathered}$$

The n- rectangle left sum for f on $\left[0,4\right]$ is

$$\begin{gathered} \sum _{i=1}^{n}f\left(x_i\right)∆x=\underset{i=1}{\overset{n}{\sum f\left(\frac{4i}{n}\right)∆x}} \\ =\underset{i=1}{\overset{n}{\sum {\left(3\left(\frac{4i}{n}\right)+2\right)}^2}}\left(\frac{1}{n}\right) \\ =\frac{144}{n^3}\sum _{i=1}^n{i}^2+\frac{48}{n^2}\sum _{i=1}^{n}i+\frac{4}{n}\underset{i=1}{\overset{n}{\sum 1}} \\ =\frac{144}{n^3}\times \frac{n(n+1)(2n+1)}{6}+\frac{48}{n^2}\frac{n(n+1)}{2}+\frac{4}{n}\times n \end{gathered}$$

Take the limit of the Riemann sum,

$\underset{n\to \infty }{\lim }\left(\frac{144}{n^3}\times \frac{n(n+1)(2n+1)}{6}+\frac{48}{n^2}\frac{n(n+1)}{2}+\frac{4}{n}\times n\right)=76$

Therefore, 

${\int }_0^4{\left(3x+2\right)}^{2}dx=76$

Part (b) Step 1. Given information

The given integral ${\int }_0^4{\left(3x+2\right)}^{2}dx$

Part (b) Step 2. The definite integral formulas from Theorem 4.13

For any real numbers a,b and c

$$\begin{gathered} {\int }_a^b{x}^{2}dx=\frac{1}{3}\left(b^3-a^3\right) \\ {\int }_a^b{x}^{2}dx=\frac{1}{2}\left(b^2-a^2\right) \\ {\int }_a^b c dx=c(b-a) \end{gathered}$$

Consider ${\int }_0^4{(3x+2)}^{2}dx$

Here,$a=0,b=4$

$$\begin{gathered} {\int }_0^4{(3x+2)}^{2}dx=9{\int }_0^4{x}^{2}dx+12{\int }_0^{4}x dx+4{\int }_0^{4}dx \\ =9\left[\frac{1}{3}\left(4^3-0^3\right)\right]+12\left[\frac{1}{2}\left(4^2-0^2\right)\right]+4(4-0) \\ =304 \end{gathered}$$

Part (c) Step 1. Given information

The given integral ${\int }_0^4{(3x+2)}^{2}dx$

Part (c) Step 2. The Fundamental Theorem of Calculus

Consider $f\left(x\right)$ is continuous on the interval $\left[a,b\right]$.If $F\left(x\right)$ is any antiderivative of $f\left(x\right)$,then

${\int }_a^{b}f\left(x\right) dx=F\left(b\right)-F\left(a\right)$

By the Fundamental theorem of Calculus,

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right) \\ Take,f\left(x\right)={(3x+2)}^2 \end{gathered}$$

Antiderivatives of $f\left(x\right)$

$$\begin{gathered} F\left(x\right)=\frac{{\left(3x+2\right)}^3}{9}+c,where c-cons\tan t \\ {\int }_0^4{(3x+2)}^{2}dx=\left(\frac{{\left(3\times 4+2\right)}^3}{9}-\frac{{\left(3\times 0+2\right)}^3}{9}\right) \\ =304 \end{gathered}$$