Find as the limit of sums.

Find $\int_{1}^{3}\left (x ^{2} +2+e^{2x}\right )dx$ as the limit of sums.

Answers (1)

(A) $\int_{a}^{b}f\left ( x \right )dx= \underset{h\rightarrow 0}{lim\; h}\left [ f\left ( a \right )+f\left ( a+h \right )+f\left ( a+2h \right ) +\cdots f\left ( a+\left ( n-1 \right )h \right )\right ]$ where $h= \frac{b-a}{h}$
(i) Here, $a= 1$ and $b= 3$
$f\left ( x \right )= x^{2}+2+e^{2x}$
$\therefore h= \frac{3-1}{n}= \frac{2}{n}\Rightarrow nh= 2$
Now,
$I= \int_{1}^{3}\left ( x^{2}+2+e^{2x} \right )dx$
$I=\underset{h\rightarrow 0}{lim}\: h\left [ f\left ( 1 \right )+f\left ( 1+n \right ) +f\left ( 1+2h \right )\cdots f\left ( 1+\left ( n-1 \right )h \right )\right ]$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ \left ( 1+2+e^{2} \right ) +\left [ \left ( 1+A^{2}+2+e^{2\left ( 1+n \right )} \right ) \right ]+\left [ \left ( 1+2n \right ) ^{2}+2+e^{2\left ( 1+2h \right )+\cdots }\right ]\left [ \left ( 1+\left ( n-1 \right )h \right )^{2}+2+e^{2\left ( 1+\left ( n-1 \right )^{h} \right )} \right ]\right ]$ $I=\underset{h\rightarrow 0}{lim}\, h\left [ 1+2+e^{2} +1+h^{2}+2n+2+e^{2\left ( 1+n \right )}+1+4n^{2}+4n+2e^{2\left ( 1+2h \right )}+\cdots \left [ 1+\left ( n-1 \right )^{2} h^{2}+2\left ( n-1 \right )h+2+e^{2}\left ( 1+\left ( n-1 \right ) h\right )\right ]\right ]$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ 1+1+1+\cdots +2+2+\cdots 2n+4n+2\left ( n-1 \right )h+\cdots n^{2}+4n^{2} +\left ( n-1 \right )^{2}h2+\cdots +e^{2}+e^{2\left ( 1+n \right )}+e^{2\left ( 1+2h \right )}+\cdots \right ]$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ 1+1+1+\cdots 2+2+\cdots 2n+4n+2\left ( n-1 \right )h+n^{2}+4n^{2}+\left ( n-1 \right )^{2}h^{2}+\cdots +e^{2}\left ( 1+e^{2n}+e^{4n} +\cdots e^{2(n-1)h}\right ) \right ]$
$\left [ \because 1+e^{2n}+e^{4n}+\cdots e^{2\left ( n-1 \right )h}= \frac{1\left ( 1-e^{2hn} \right )}{1-e^{2n}} \right ]$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ 3n+2h\frac{n\left ( n-1 \right )}{2} +\frac{n^{2}\left ( n-1 \right )n\left ( 2n-1 \right )}{6}+\frac{e^{2}\left ( 1-e^{2hn} \right )}{1-e^{2n}}\right ]$
$\left [ \because 1+2+3+4+\cdots n= \frac{n\left ( n+1 \right )}{2} 1^{2}+2^{2} +3^{2}+n^{2}= \frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6}\right ]$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ 3nh+nh^{2}\left ( n-1 \right ) +\frac{h^{3}\left ( n-1 \right )n\left ( 2n-1 \right )}{6}+\frac{e^{2h}\left ( 1+e^{2nh} \right )}{1-e^{2n}}\right ]$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ nh\times 3+nh\times h\left ( n-1 \right )+\frac{nhh^{2}\left (n-1 \right )\left ( n-1 \right )}{6}+\frac{e^{2}h\left ( 1+e^{2nh} \right )}{1-e^{2n}}\right ]$
put $nh= 2$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ 6+2+\left ( 2-n \right )+\frac{\left ( nh-n \right )nh\left ( 2nh-h \right )}{6} +\frac{e^{2}}{2}\times \frac{2h\left ( 1-e^{4} \right )}{1-e^{2n}}\right ]$
$I=\underset{h\rightarrow 0}{lim}\, h\left [ 6+2+\left ( 2-n \right )+\frac{\left ( 2-n \right )2\left ( 4-h \right )}{6}+\frac{e^{2}}{2} \left ( -1 \right )\left ( 1-e^{4} \right )\right ]$
Now,
$I= 6+4+\frac{2\times 2\times 4}{6}-\frac{e^{2}\left ( 1-e^{4} \right )}{2}$
$\Rightarrow I= 10+\frac{8}{3}-\frac{e^{2}}{2}\left ( 1-e^{4} \right )$
$\Rightarrow I= \frac{38}{3}-\frac{e^{2}}{2}\left ( 1-e^{4} \right )$