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Evaluate the definite integral as limits of sums\[\int\limits_0^4(x+e^{2x})dx\]

$\begin{array}{1 1}8+\large\frac{e^8-1}{2} \\7+\large\frac{e^8+1}{2} \\6+\large\frac{e^8}{2} \\5+\large\frac{e^8-3}{2}\end{array} $

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  • $\int\limits_a^b f(x)dx=lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h]$
  • where $ h=\large\frac{b-a}{n} $
  • (ii) $ \lim_{h \to 0} \Large\frac{e^{nh}-1}{e^h-1}=1$
Given $ I= \int \limits_0^4 (x+e^{2x})dx$
$\int \limits_a^b f(x)dx=lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h]$
Here $ f(x)=x+e^{2x},\;a=0,\;\;b=4\:and\:nh=b-a=4$
Hence $\int \limits_0^4 (x+e^{2x})dx=\lim_ {h \to 0} h[f(0)+f(0+h)+...f(0+(n-1)h]$
Now substituting the above values we get.
$\int \limits_0^4 (x+e^{2x})dx=\lim_ {h \to 0} h[(0+h)+e^{2(0+h)}+(0+2h)+e^{(0+2h)}+......(0+(n-1)h+e^{0+h-1)}h]$
Seperating the terms
$=\lim_{h \to 0}h[h+2h+3h+.....(n-1)h+e.e^h+e^0.e^{4h}+.....e^0.e^{2(n-1)h}]$
$=\lim_{h \to 0}h[h(1+2+3+.....(n-1)]+[1+e^{2h}+e^{4h}+.....e^{2(n-1)h}]$
But we know $1+2+.......(n-1)=\frac {n(n-1)}{2}$
Since $1+e^{2h}+e^{4h}+.......e^{2(n-1)h}$ is in form a geometric progression
we know $a+ar+ar^2+..........ar{n-1}=a \bigg(\large\frac{r^n-1}{r-1}\bigg)$
here,$ n^{(2nh)}\;r=e\;and\;a=1$
$\lim_{h \to \infty} \large\frac{e^h-1}{h}=1$
Therefore, $ 1+e^{2h}+e^{4h}+.......e^{2(n-1)h}$
Substituting this in I we get,
$I=\lim _{h \to 0} \bigg\{\bigg[\large\frac{h(n)(nh-h)}{2}\bigg]+h\bigg[\large\frac{e^{2(nh)}-1}{e^{2h}-1}\bigg]\bigg\}$
$=\lim _{h \to 0} \bigg\{\bigg[\large\frac{4(4-h)}{2}\bigg]+h\bigg[\large\frac{e^{8}-1}{e^{2h}-1}\bigg]\bigg\}$
On applying limits,
$\large\frac{ 4 \times 4}{2}+\Large\frac{e^8-1}{2}$$=8+\large\frac{e^8-1}{2}$
answered Feb 10, 2013 by meena.p
edited Dec 19, 2013 by rvidyagovindarajan_1
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