Question
Evalaute : $\int_1^2(2 x+5) d x$
✓
Answer
$
\begin{aligned}
& \text { Given, } \int_1^2(2 x+5) d x=\int_a^b f(x) d x \\
& f(x)=2 x+5 \quad a=1 ; b=2 \\
& \Rightarrow \quad f(a+r h)=f(1+r h) \quad \text { and } \quad h=\frac{b-a}{n} \\
& =2(1+r h)+5 \\
& h=\frac{2-1}{n} \\
& =2+2 r h+5 \\
& =7+2 r h \\
& \therefore \quad n h=1 \\
\end{aligned}
$
We know $\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot f(a+r h)$
$\begin{aligned} \therefore \quad \int_1^2(2 x+5) d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot(7+2 r h) \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(7 h+2 r h^2\right) \\ & =\lim _{n \rightarrow \infty}\left(7 h \sum_{r=1}^n 1+2 h^2 \sum_{r=1}^n r\right) \\ & =\lim _{n \rightarrow \infty}\left[7 h \cdot(n)+2 h^2\left(\frac{n(n+1)}{2}\right)\right] \\ & =\lim _{n \rightarrow \infty}\left[7 n h+h^2 n^2\left(1+\frac{1}{n}\right)\right] \\ & =\lim _{n \rightarrow \infty}\left[7(1)+(1)^2\left(1+\frac{1}{n}\right)\right] \\ & =7+1(1+0)=8\end{aligned}$
\begin{aligned}
& \text { Given, } \int_1^2(2 x+5) d x=\int_a^b f(x) d x \\
& f(x)=2 x+5 \quad a=1 ; b=2 \\
& \Rightarrow \quad f(a+r h)=f(1+r h) \quad \text { and } \quad h=\frac{b-a}{n} \\
& =2(1+r h)+5 \\
& h=\frac{2-1}{n} \\
& =2+2 r h+5 \\
& =7+2 r h \\
& \therefore \quad n h=1 \\
\end{aligned}
$
We know $\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot f(a+r h)$
$\begin{aligned} \therefore \quad \int_1^2(2 x+5) d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot(7+2 r h) \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(7 h+2 r h^2\right) \\ & =\lim _{n \rightarrow \infty}\left(7 h \sum_{r=1}^n 1+2 h^2 \sum_{r=1}^n r\right) \\ & =\lim _{n \rightarrow \infty}\left[7 h \cdot(n)+2 h^2\left(\frac{n(n+1)}{2}\right)\right] \\ & =\lim _{n \rightarrow \infty}\left[7 n h+h^2 n^2\left(1+\frac{1}{n}\right)\right] \\ & =\lim _{n \rightarrow \infty}\left[7(1)+(1)^2\left(1+\frac{1}{n}\right)\right] \\ & =7+1(1+0)=8\end{aligned}$
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