# Use Definition 2 to find an expression for the area under the graph of f as a li

Use Definition 2 to find an expression for the area under the graph of f as a limit. Do not evaluate the limit. $f\left(x\right)={x}^{2}+{\left(1+2x\right)}^{\frac{1}{2}},4\le x\le 7$
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Step1
We know that,
$Area={\int }_{a}^{b}f\left(x\right)dx=\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Delta }x\sum _{i=1}^{n}f\left(a+i\mathrm{\Delta }x\right)$
Where $\mathrm{\Delta }x=\frac{b-a}{n}$
Step 2
We have
$Area={\int }_{4}^{7}{x}^{2}+\sqrt{1+2x}dx$
 In this problem $f\left(x\right)={x}^{2}+\sqrt{1+2x}a=4,b=7$ Hence $\mathrm{\Delta }x=\frac{7-4}{n}=\frac{3}{n}$
$Area=\underset{n\to \mathrm{\infty }}{lim}\frac{3}{n}\sum _{i=1}^{n}f\left(4+i\cdot \frac{3}{n}\right)$
$Area=\underset{n\to \mathrm{\infty }}{lim}\frac{3}{n}\sum _{i=1}^{n}\left[{\left(4+i\cdot \frac{3}{n}\right)}^{2}+\sqrt{1+2\left(4+i\cdot \frac{3}{n}\right)}\right]$
$Area=\underset{n\to \mathrm{\infty }}{lim}\frac{3}{n}\sum _{i=1}^{n}\left[{\left(4+\frac{3i}{n}\right)}^{2}+\sqrt{9+\frac{6i}{n}}\right]$
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