Use the Definition of Area as a Limit to Find$f ( x ) = 2 x$

Question

Examlex · Accepted Answer

The Answer is Width:

\Delta x = \frac { 3 - 1 } { n } = \frac { 2 } { n }

, REP:

x _ { k } = ( 1 ) + k \left( \frac { 2 } { n } ight) = 1 + \frac { 2 k } { n }

, Height:

f \left( x _ { k } ight) = 2 \left( 1 + \frac { 2 k } { n } ight) = 2 + \frac { 4 k } { n }

.

\begin{aligned}A & = \lim _ { n ightarrow \infty } \sum _ { k = 1 } ^ { n } f \left( x _ { k } ight) \Delta x = \lim _ { n ightarrow \infty } \sum _ { k = 1 } ^ { n } \left[ \left( 2 + \frac { 4 k } { n } ight) \cdot \frac { 2 } { n } ight] = \lim _ { n ightarrow \infty } \sum _ { k = 1 } ^ { n } \left( \frac { 4 } { n } + \frac { 8 k } { n ^ { 2 } } ight) \& = \lim _ { n ightarrow \infty } \frac { 4 } { n } \sum _ { k = 1 } ^ { n } 1 + \lim _ { n ightarrow \infty } \frac { 8 } { n ^ { 2 } } \sum _ { k = 1 } ^ { n } k = \lim _ { n ightarrow \infty } \left[ \frac { 4 } { n } \cdot n ight] + \lim _ { n ightarrow \infty } \left[ \frac { 8 } { n ^ { 2 } } \cdot \frac { n ( n + 1 ) } { 2 } ight] = \lim _ { n ightarrow \infty } [ 4 ] + \lim _ { n ightarrow \infty } \left[ 4 + \frac { 4 } { n } ight] = 8\end{aligned}

Use the Definition of Area as a Limit to Find $f ( x ) = 2 x$