Evaluate the following definite integral by the limit definition.

Question

Quizplus · Accepted Answer

The limit definition of the definite integral is:
$$\int_a^b f(x) dx = \lim_{n 	o \infty} \sum_{i=1}^n f(x_i^*) \Delta x,$$
where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is any point in the $i$th subinterval $[x_{i-1},x_i]$. 
In this case, we have
$$\int_0^2 (2x-1) dx = \lim_{n 	o \infty} \sum_{i=1}^n f(x_i^*) \Delta x,$$
where $f(x) = 2x-1$, $a=0$, $b=2$, $\Delta x = \frac{2-0}{n} = \frac{2}{n}$, and $x_i^* = 0 + i \cdot \frac{2}{n}$. 
The sum is
\begin{align*}
\sum_{i=1}^n f(x_i^*) \Delta x &= \sum_{i=1}^n \left(2x_i^* - 1ight) \cdot \frac{2}{n} \
&= \sum_{i=1}^n \left(2 \cdot \frac{2i}{n} - 1ight) \cdot \frac{2}{n} \
&= \frac{4}{n} \sum_{i=1}^n \frac{i}{n} - \frac{2}{n} \sum_{i=1}^n 1 \
&= \frac{4}{n^2} \sum_{i=1}^n i - \frac{2}{n} \cdot n \
&= \frac{4}{n^2} \cdot \frac{n(n+1)}{2} - 2 \
&= \frac{2n+4}{n} - 2 \
&= \frac{2n+4-2n}{n} \
&= \frac{4}{n}.
\end{align*}
Thus, we have
$$\int_0^2 (2x-1) dx = \lim_{n 	o \infty} \frac{4}{n} = \boxed{	extbf{(B) }0}.$$

Evaluate the Following Definite Integral by the Limit Definition