Express the limit as a deﬁnite integral.
- $\text { where } P \text { is a partition of }[2,4]$
A) $\int _ { 1 } ^ { \mathrm { n } } \frac { 2 } { 3 - 6 \mathrm { x } } \mathrm { dx }$
B) $\int _ { 2 } ^ { 4 } \frac { 2 } { 3 - 6 x } d x$
C) $\int _ { 4 } ^ { 2 } \frac { 2 } { 3 - 6 x ^ { 2 } } d x$
D) $\int _ { 2 } ^ { 4 } \frac { 2 } { 3 - 6 x ^ { 2 } } d x$

Question

Express the limit as a deﬁnite integral.
-  $\text { where } P \text { is a partition of }[2,4]$ 
A) $\int _ { 1 } ^ { \mathrm { n } } \frac { 2 } { 3 - 6 \mathrm { x } } \mathrm { dx }$ 
B) $\int _ { 2 } ^ { 4 } \frac { 2 } { 3 - 6 x } d x$ 
C) $\int _ { 4 } ^ { 2 } \frac { 2 } { 3 - 6 x ^ { 2 } } d x$ 
D) $\int _ { 2 } ^ { 4 } \frac { 2 } { 3 - 6 x ^ { 2 } } d x$

Quizplus · Accepted Answer

The limit can be written as:
$$\lim_{\Vert P\Vertightarrow 0}\sum_{i=1}^n f(c_i)\Delta x_i$$
where $f(x) = \frac{2}{3-6x^2}$ and $P$ is a partition of $[2,4]$. We can choose $c_i$ to be any point in the $i$th subinterval of $P$. One convenient choice is the midpoint $c_i = \frac{x_{i-1}+x_i}{2}$.
Then, we have
\begin{align*}
\lim_{\Vert P\Vertightarrow 0}\sum_{i=1}^n f(c_i)\Delta x_i &= \lim_{\Vert P\Vertightarrow 0}\sum_{i=1}^n \frac{2}{3-6c_i^2}\Delta x_i\
&= \int_2^4 \frac{2}{3-6x^2}dx
\end{align*}
Therefore, the answer is D, with the integral $\int_{2}^{4} \frac{2}{3-6x^{2}} dx$.

Express the Limit as a Deﬁnite Integral where P is a partition of [2,4]\text { where } P \text { is a partition of }[2,4] where P is a partition of [2,4]

Express the Limit as a Deﬁnite Integral $\text { where } P \text { is a partition of }[2,4]$