Compute the deﬁnite integral as the limit of Riemann sums.
-$$\int _ { 1 } ^ { 0 } ( 3 x - 1 ) d x$$
A) $- \frac { 5 } { 2 }$
B) $\frac { 1 } { 2 }$
C) 0
D) $- \frac { 1 } { 2 }$

Question

Quizplus · Accepted Answer

We have 
\begin{align*}
\int _ { 1 } ^ { 0 } ( 3 x - 1 ) d x &= \lim_{n	o \infty} \sum_{i=1}^n f(c_i)\Delta x\
&= \lim_{n	o \infty} \sum_{i=1}^n (3c_i-1)\frac1n\
\end{align*}
where $f(x)=3x-1$ and $\Delta x=\frac1n$. 
The $c_i$'s are $1/n,2/n,\ldots,n/n$. We choose the right endpoints for convenience. 
Thus, 
\begin{align*}
\int _ { 1 } ^ { 0 } ( 3 x - 1 ) d x &= \lim_{n	o \infty} \sum_{i=1}^n (3c_i-1)\frac1n\
&= \lim_{n	o \infty} \sum_{i=1}^n \left(3\cdot \frac in-1ight)\frac1n\
&= \lim_{n	o \infty} \frac1n\sum_{i=1}^n \left(\frac{3i}n-1ight)\
&= \lim_{n	o \infty} \frac1n\left(3\frac1n+3\frac2n+3\frac3n+\cdots+3\frac nn - night)\
&= \lim_{n	o \infty} \left(\frac{3}{n^2}\sum_{i=1}^n i - 1ight)\
&= \lim_{n	o \infty} \left(\frac{3}{n^2}\cdot \frac{n(n+1)}{2} - 1ight)\
&= \lim_{n	o \infty} \left(\frac{3n+3}{2n} - 1ight)\
&= \frac32 -1\
&= -\frac12
\end{align*}

Compute the Deﬁnite Integral as the Limit of Riemann Sums ∫10(3x−1)dx\int _ { 1 } ^ { 0 } ( 3 x - 1 ) d x∫10​(3x−1)dx

Compute the Deﬁnite Integral as the Limit of Riemann Sums $\int _ { 1 } ^ { 0 } ( 3 x - 1 ) d x$