The expression for the area under the graph is [tex]A = \lim_{n \to \infty} \left(\frac{2}{n}\right)\Sigma_{i=1}^{n} \left(4+\frac{2\cdot i}{n} \right)^{2} + \lim_{n \to \infty} \left(\frac{2}{n} \right)\Sigma_{i = 1}^{n}\sqrt{1+2\cdot \left(4+\frac{2\cdot i}{n} \right)}[/tex].
In this case, we must assume that the area below the curve is equivalent to the sum of infinite rectangles, that is to say:
[tex]A = \lim_{n \to \infty} \left(\frac{b-a}{n}\right) \Sigma_{i=1}^{n} f\left[a + i\cdot \left(\frac{b-a}{n} \right)\right][/tex] (1)
If we know that [tex]a = 4[/tex], [tex]b =6[/tex] and [tex]f(x) = x^{2}+\sqrt{1+2\cdot x}[/tex], then we have the following expression:
[tex]A = \lim_{n \to \infty} \left(\frac{2}{n} \right)\Sigma_{i=1}^{n} f\left(4+\frac{2\cdot i}{n} \right)[/tex]
[tex]A = \lim_{n \to \infty} \left(\frac{2}{n} \right)\Sigma_{i=1}^{n}\left[\left(4+\frac{2\cdot i}{n} \right)^{2}+\sqrt{1+2\cdot \left(4+\frac{2\cdot i}{n} \right)}\right][/tex]
[tex]A = \lim_{n \to \infty} \left(\frac{2}{n}\right)\Sigma_{i=1}^{n} \left(4+\frac{2\cdot i}{n} \right)^{2} + \lim_{n \to \infty} \left(\frac{2}{n} \right)\Sigma_{i = 1}^{n}\sqrt{1+2\cdot \left(4+\frac{2\cdot i}{n} \right)}[/tex]
The expression for the area under the graph is [tex]A = \lim_{n \to \infty} \left(\frac{2}{n}\right)\Sigma_{i=1}^{n} \left(4+\frac{2\cdot i}{n} \right)^{2} + \lim_{n \to \infty} \left(\frac{2}{n} \right)\Sigma_{i = 1}^{n}\sqrt{1+2\cdot \left(4+\frac{2\cdot i}{n} \right)}[/tex].
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Nota - The statement has some typing errors. Corrected statement is presented below:
Use the definition to find an expression for the area under the graph of [tex]f[/tex] as a limit. Do not evaluate the limit: [tex]f(x) = x^{2}+\sqrt{1+2\cdot x}[/tex], [tex]4\le x \le 6[/tex].
