Find the formula and limit as requested.
-For the function $f ( x ) = 12 - 2 x ^ { 2 }$, find a formula for the lower sum obtained by dividing the interval $[ 0,1 ]$ into $n$ equal subintervals. Then take the limit as $n \rightarrow \infty$ to calculate the area under the curve over $[ 0,1 ]$.
A) $12 - \frac { 4 n ^ { 3 } + 6 n ^ { 2 } + 2 n } { 3 n ^ { 3 } }$; Area $= \frac { 34 } { 3 }$
B) $\frac { 2 n ^ { 3 } + 3 n ^ { 2 } + n } { 3 n ^ { 3 } } ;$ Area $= \frac { 2 } { 3 }$
C) $12 - \frac { 2 n ^ { 3 } + 3 n ^ { 2 } + n } { 3 n ^ { 3 } } ;$ Area $= \frac { 34 } { 3 }$
D) $12 + \frac { 2 n ^ { 3 } + 3 n ^ { 2 } + 1 } { 3 n ^ { 3 } } ;$ Area $= \frac { 38 } { 3 }$

Question

Quizplus · Accepted Answer

To find the lower sum, we need to take the minimum value of the function in each subinterval and multiply it by the width of the subinterval.

The width of each subinterval is $\frac{1}{n}$. To find the minimum value of the function in each subinterval, we can evaluate the function at the endpoints and the midpoint, and take the minimum of those three values.

Let $x_i$ be the left endpoint of the $i$th subinterval, $x_{i+\frac{1}{2}}$ be the midpoint of the $i$th subinterval, and $x_{i+1}$ be the right endpoint of the $i$th subinterval. Then the lower sum is:

\begin{align*}
L_n &= \sum_{i=0}^{n-1} \left(12 - 2\left(x_{i+\frac{1}{2}}ight)^2ight)\left(\frac{1}{n}ight) \
&= \sum_{i=0}^{n-1} 12\left(\frac{1}{n}ight) - 2\left(\frac{1}{n}ight)\left(x_{i+\frac{1}{2}}^2ight) \
&= 12 - \frac{2}{n}\sum_{i=0}^{n-1} x_{i+\frac{1}{2}}^2 \
&= 12 - \frac{2}{n}\sum_{i=0}^{n-1} \left(\frac{i+1/2}{n}ight)^2 \
&= 12 - \frac{2}{n^3}\sum_{i=1}^{n} i^2 \
&= 12 - \frac{2}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} \
&= 12 - \frac{2n^2 + 3n + 1}{3n^2} \
&= 12 - \frac{2}{3}\cdot\frac{n^2 + \frac{3}{2}n + \frac{1}{2}}{n^2} \
&= 12 - \frac{2}{3}\left(\frac{n}{n}ight)^2\left(1+\frac{3}{2n}+\frac{1}{2n^2}ight) \
&= 12 - \frac{2}{3}\left(1 + \frac{3}{2n} + \frac{1}{2n^2}ight).
\end{align*}

Taking the limit as $nightarrow\infty$, we get:

\begin{align*}
\lim_{nightarrow\infty} L_n &= \lim_{nightarrow\infty}\left(12 - \frac{2}{3}\left(1 + \frac{3}{2n} + \frac{1}{2n^2}ight)ight) \
&= 12 - \frac{2}{3}\cdot 1 \
&= \frac{34}{3}.
\end{align*}

Therefore, the area under the curve is $\frac{34}{3}$, and the best choice is $\boxed{	ext{(C)}}$.

Find the Formula and Limit as Requested f(x)=12−2x2f ( x ) = 12 - 2 x ^ { 2 }f(x)=12−2x2

Find the Formula and Limit as Requested $f ( x ) = 12 - 2 x ^ { 2 }$