91Ó°ÊÓ

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

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 fon $\left[0,4\right]$is

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

Take the limit of the Riemann sum,

$\underset{n→∞}{\lim }\left(\frac{144}{n^3}×\frac{n(n+1)(2n+1)}{6}+\frac{48}{n^2}\frac{n(n+1)}{2}+\frac{4}{n}×n\right)=76$

Therefore,

${∫}_0^4{\left(3x+2\right)}^{2}dx=76$

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

For any real numbers a,band c

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

Consider role="math" localid="1649924142154" ${∫}_0^4{(3x+2)}^{2}dx$

Here,$a=0,b=4$

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

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

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

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

By the Fundamental theorem of Calculus,

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

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

role="math" localid="1649924770198" $$\begin{gathered} F\left(x\right)=\frac{{\left(3x+2\right)}^3}{9}+c,wherec-cons\tan t \\ {∫}_0^4{(3x+2)}^{2}dx=\left(\frac{{\left(3×4+2\right)}^3}{9}-\frac{{\left(3×0+2\right)}^3}{9}\right) \\ =304 \end{gathered}$$