AP Calculus AB / Integration and Accumulation of Change / Question No. 8

8. The expression $\frac{1}{20} [{(\frac{1}{20})}^{2} + {(\frac{2}{20})}^{2} + {(\frac{3}{20})}^{2} + \dots + {(\frac{20}{20})}^{2}]$ is a Riemann sum approximation for

\frac{1}{20} \int_{0}^{20} x^{2} d x

\frac{1}{20} \int_{0}^{1} x^{2} d x

\int_{0}^{1} x^{2} d x

\int_{0}^{1} \frac{1}{x^{2}} d x

Answer is: option3

\int_{0}^{1} x^{2} d x

Solution:

The given expression can be written as:

$\sum_{k = 1}^{20} (\frac{1}{20}) {(\frac{k}{20})}^{2}$

This matches the general form of a Riemann sum:

$\sum_{k = 1}^{n} (\frac{b - a}{n}) f (a + \frac{(b - a) k}{n})$

where:

$n = 20$ (number of subintervals),
$\frac{b - a}{n} = \frac{1}{20}$ (width of each subinterval),
$f (a + \frac{(b - a) k}{n}) = {(\frac{k}{20})}^{2}$ (function evaluated at sample points).

From $\frac{b - a}{20} = \frac{1}{20}$ , we get $b - a = 1$ .

The sample points are $\frac{k}{20}$ , which suggests $a = 0$ (since when $k = 1$ , the first point is $\frac{1}{20}$ ).

Thus, $b = 1$ (since $b - a = 1$ and $a = 0$ ).

The function evaluated at the sample points is:

$f (\frac{k}{20}) = {(\frac{k}{20})}^{2} \Rightarrow f (x) = x^{2}$

The Riemann sum approximates:

$\int_{a}^{b} f (x) d x = \int_{0}^{1} x^{2} d x$

This corresponds to option (C).

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