Compute the deﬁnite integral as the limit of Riemann sums.
-$$\int _ { p } ^ { q } 4 x ^ { 2 } , a

Question

Compute the deﬁnite integral as the limit of Riemann sums.
-$$\int _ { p } ^ { q } 4 x ^ { 2 } , a < b$$
A) $q ^ { 2 } - p ^ { 2 }$
B) $64 \left( \frac { q ^ { 3 } } { 3 } - \frac { p ^ { 3 } } { 3 } \right)$
C) $4 \left( \frac { q ^ { 3 } } { 3 } - \frac { p ^ { 3 } } { 3 } \right)$
D) $16 ( q - p )$

Quizplus · Accepted Answer

The Riemann sum for the given integral is:
$$\sum_{i=1}^{n}4x_i^2\Delta x$$
where $\Delta x = \frac{q-p}{n}$ is the width of each subinterval and $x_i$ is any point in the $i^{th}$ subinterval.
Now, we can choose $x_i$ to be the right endpoint of each subinterval, i.e. $x_i=p+i\Delta x$ for $i=1,2,\ldots,n$. Then, the Riemann sum simplifies as:
\begin{align*}
\sum_{i=1}^{n}4x_i^2\Delta x &= \sum_{i=1}^{n}4(p+i\Delta x)^2\Delta x\
&=4\Delta x\sum_{i=1}^{n}(p+i\Delta x)^2\
&=4\Delta x\sum_{i=1}^{n}\left(p^2+2p\Delta x i + (\Delta x i)^2 ight)\
&=4\Delta x\left(np^2+2p\Delta x\sum_{i=1}^{n}i + \Delta x^2\sum_{i=1}^{n}i^2ight)\
&=4\Delta x\left(np^2+2p\Delta x\frac{n(n+1)}{2} + \Delta x^2\frac{n(n+1)(2n+1)}{6}ight)\
&=4\Delta x\left(np^2+p\Delta x n(n+1) + \Delta x^2\frac{n(n+1)(2n+1)}{3}ight)\
&=4\frac{q-p}{n}\left(np^2+p\frac{q-p}{n} n(n+1) + \frac{(q-p)^2}{n^2}\frac{n(n+1)(2n+1)}{3}ight)\
&=4(q-p)\left(p+\frac{q-p}{n} \frac{n+1}{2} + \frac{(q-p)^2}{n^2}\frac{(n+1)(2n+1)}{3}ight)\
&=4(q-p)\left(\frac{2p+q-p}{2}+\frac{(q-p)^2}{3n}+\frac{(q-p)^3}{3n^2}\frac{2n+1}{3}ight)\
&=4(q-p)\left(\frac{q+p}{2}+\frac{(q-p)^2}{3n}+\frac{(q-p)^3}{9n^2}(2n+1)ight)\
\end{align*}
As $n	o\infty$, we have $\Delta x	o 0$ and $np	o (q-p)p=a$ and $n(q-p)	o (q-p)q-b$. Hence, we have:
\begin{align*}
\lim_{n	o\infty}\sum_{i=1}^{n}4x_i^2\Delta x &= 4(q-p)\left(\frac{q+p}{2}ight)+4\left(\frac{q-p}{3}ight)^2(q-p)+4\left(\frac{q-p}{3}ight)^3\left(q+b-2pight)\
&=4\left(\frac{q-p}{3}ight)^3\left(9(q-p)+3(q+b-2p)+12(q-p)ight)\
&=4\left(\frac{q-p}{3}ight)^3\left(22(q-p)+3bight)\
&=4\cdot\frac{(q-p)^3}{27}\cdot\left(22q-22p+3bight)
\end{align*}
Thus, we have:
$$\int_{p}^{q}4x^2dx=\lim_{n	o\infty}\sum_{i=1}^{n}4x_i^2\Delta x = 4\cdot\frac{(q-p)^3}{27}\cdot\left(22q-22p+3bight)$$
Comparing this with the given options, we see that the correct choice is $\boxed{	extbf{(C)}}$.

Compute the Deﬁnite Integral as the Limit of Riemann Sums ∫pq4x2,a<b\int _ { p } ^ { q } 4 x ^ { 2 } , a < b∫pq​4x2,a<b

Compute the Deﬁnite Integral as the Limit of Riemann Sums $\int _ { p } ^ { q } 4 x ^ { 2 } , a < b$