Question
Evalaute : $\int_2^3 7^x \cdot d x$

Answer

$
\begin{aligned}
& \text { Given, } \int_2^3 7^x \cdot d x=\int_a^b f(x) d x \\
& f(x)=7^x \quad a=2 ; b=3 \\
& \Rightarrow \quad f(a+r h)=f(1+r h) \\
& =7^{2+r h} \\
& =7^2 \cdot 7^{\text {th }} \\
& h=\frac{b-a}{n} \\
& h=\frac{3-2}{n} \\
& \therefore \quad n h=1 \\
\end{aligned}
$
$
\therefore \quad n h=1
$
We know $\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot f(a+r h)$
$
\therefore \quad \begin{aligned}
\int_1^3 7^x \cdot d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot\left(7^2 \cdot 7^{r h}\right) \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot \sum_{r=1}^n h \cdot 7^{r \cdot h} \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot h \cdot\left[7^h+7^{2 h}+7^{3 h}+7^{4 h}+\ldots+7^{h h}\right] \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot h \cdot\left(\frac{7^h\left[\left(7^h\right)^n-1\right]}{7^h-1}\right)=\lim _{n \rightarrow \infty} 7^2 \cdot\left(\frac{7^h\left(7^{n h}-1\right)}{\left.\frac{7^h-1}{h}\right)}\right. \\
& =\lim _{n \rightarrow \infty} 7^2 \cdot\left(\frac{7^h\left(7^{(1)}-1\right)}{\left.\frac{7^h-1}{h}\right)}\right. \\
& =\frac{7^2 \cdot 7^0 \cdot(7-1)}{\log 7}=\frac{(49)(1)(6)}{\log 7}=\frac{294}{\log 7}
\end{aligned}
$

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