Riemann Sums

## **Problem**

**Part 1:** Convert the following Riemann sum as a definite integral and evaluate the definite integral.

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

{\displaystyle \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) }

**Part 2:** Use the finite sums formulas to calculate the Riemann sum.

## **Solution**

**Part 1**

For the above Riemann sum

x
i

=
a

(
b
−
a
)
i

n

=
1

2
i

n

{\displaystyle x\_i = a + \frac{(b-a)i}{n} = 1 + \frac{2i}{n} }

. By matching terms, we find

# a
1

{\displaystyle a = 1}

 and

b
−
# a
2

{\displaystyle b - a = 2 }

. So

# b
3

{\displaystyle b=3}

. Subsequently,

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

=

∫

1

3

x
2

−
2
x
1

d
x

{\displaystyle \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) = \int\_{1}^{3} x^2-2x + 1 \,dx }

∫

1

3

x
2

−
2
x
1

d
x

=

x
3

3

−

x
2

x

1
3

=

8
3

.

{\displaystyle \begin{align}
\int\_{1}^{3} x^2-2x + 1 \, dx &= \frac{x^3}{3} - x^2 + x \,\Biggr|\_1^3 \\[5 pt]
&= \frac{8}{3}.
\end{align} }

Therefore,

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

=

8
3

.

{\displaystyle \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) = \frac{8}{3}. }

**Part 2**

First simplify the expression in the Riemann sum

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

=

lim

n
→
∞

∑

# i
1

n

[

4

i
2

n
2

]

(

2
n

)

=

lim

n
→
∞

∑

# i
1

n

[

8

i
2

n
3

]

=
8

lim

n
→
∞

[

(

∑

# i
1

n

i
2

)

1

n
3

]

{\displaystyle \begin{align}
\lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) &= \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[\frac{4i^2}{n^2} \right] \left(\frac{2}{n} \right) \\[5 pt]
 &= \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[\frac{8i^2}{n^3} \right] \\[5 pt]
 &= 8\lim\_{n \to \infty} \left[\left(\sum\_{i=1}^{n} i^2 \right) \frac{1}{n^3} \right]
 \end{align} }

Since

∑

# i
1

n

i

2

=

n
(
n
1
)
(
2
n
1
)

6

{\displaystyle \sum \_{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}}

, the Riemann sum simplifies to

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

=
8

lim

n
→
∞

[

n
(
n
1
)
(
2
n
1
)

6

1

n
3

]

{\displaystyle \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) = 8\lim\_{n \to \infty} \left[\frac{n(n+1)(2n+1)}{6} \frac{1}{n^3} \right] }

or

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

=

8
6

lim

n
→
∞

(

n
1

n

)

(

2
n
1

n

)

{\displaystyle \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) = \frac{8}{6} \lim\_{n \to \infty} \left(\frac{n+1}{n}\right) \left(\frac{2n+1}{n}\right) }

or even better

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

=

8
6

lim

n
→
∞

(

1

1
n

)

(

1

2
n

)

.

{\displaystyle \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) = \frac{8}{6} \lim\_{n \to \infty} \left(1+ \frac{1}{n}\right) \left(1+\frac{2}{n}\right). }

Therefore,

lim

n
→
∞

∑

# i
1

n

[

(

1

2
i

n

)

2

−
2

(

1

2
i

n

)

1

]

(

2
n

)

=

8
3

.

{\displaystyle \lim\_{n \to \infty} \sum\_{i=1}^{n} \left[{\left(1 + \frac{2i}{n} \right)}^{2} -
 2 \left(1+ \frac{2i}{n}\right) + 1 \right] \left(\frac{2}{n} \right) = \frac{8}{3}. }