Question

determine a region whose area is equal to the given limit. do not evaluate the limit.

$$\\lim _ { n \ ightarrow \\infty } \\sum _ { i = 1 } ^ { n } \\frac { 4 } { n } \\sqrt { 1 + \\frac { 4 i } { n } }$$

$$4 \sqrt { 1 - x }$$ on $$ 0,4 $$

$$\sqrt { 1 + x }$$ on $$ 0,4 $$

$$4 \sqrt { 1 + x }$$ on $$ - 4,4 $$

$$4 \sqrt { 1 + x }$$ on $$ 0,4 $$

$$\sqrt { 1 + x }$$ on $$ - 4,4 $$

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Explanation:

Step1: Recall the formula for the definite integral as a limit of Riemann sums

The formula for the definite integral \(\int_{a}^{b}f(x)dx=\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}f(x_{i})\Delta x\), where \(\Delta x=\frac{b - a}{n}\) and \(x_{i}=a + i\Delta x\).

In the given limit \(\lim_{n
ightarrow\infty}\sum_{i=1}^{n}\frac{4}{n}\sqrt{1+\frac{4i}{n}}\), we have \(\Delta x=\frac{4}{n}\).

Step2: Identify \(a\), \(b\) and \(f(x)\)

Since \(\Delta x=\frac{b - a}{n}=\frac{4}{n}\), then \(b - a=4\).

Also, \(x_{i}=a + i\Delta x\). Here \(x_{i}=1+\frac{4i}{n}\), if we let \(a = 0\), then \(b=4\) (because \(\Delta x=\frac{4}{n}\) and \(x_{i}=a + i\Delta x\)). And \(f(x_{i})=\sqrt{1 + x_{i}}\) (comparing with \(\sum_{i = 1}^{n}f(x_{i})\Delta x\), where \(\Delta x=\frac{4}{n}\) and \(x_{i}=\frac{4i}{n}\)). But our sum is \(\sum_{i=1}^{n}\frac{4}{n}\sqrt{1+\frac{4i}{n}}\), so \(f(x)=4\sqrt{1 + x}\) (because \(\sum_{i = 1}^{n}f(x_{i})\Delta x\) and \(\Delta x=\frac{4}{n}\), \(x_{i}=\frac{4i}{n}\) and \(f(x_{i}) = 4\sqrt{1+x_{i}}\)).

Answer:

\(4\sqrt{1 + x}\) on \([0,4]\) (i.e., the fourth option)