PRACTICE PROBLEMS: SET 1
MATH 101: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. What is the integral represented by the following limit?
( )
n
iπ
π ∑
·
tan
lim
n→∞ 4n
4n
i=1
Do not evaluate the integral.
Problem 2. Express
1∑
1
( i )2
n→∞ n
i=1 1 +
n
lim
n
as a definite integral. Do not evaluate the integral.
Problem 3. A particle moves on the x-axis with velocity given by the formula
f (t) = t2 . Find the distance traveled by the particle in the time interval t = 3 to
t = 5.
You may use the formulas:
n
∑
i=
i=1
n(n + 1)
2
and
n
∑
i2 =
i=1
Problem 4. Evaluate
∫
n(n + 1)(2n + 1)
.
6
4
−2
(x2 − 5) dx
using a limit of Riemann sums.
1
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Problem 1. We are looking for an answer of the form
∫ b
f (x) dx;
a
so, we need to determine the endpoints of integrations: a and b, and also the
function f (x).
∑
π
The factor 4n
which is infront of the above -notation represents ∆x, which
always equals b−a
n . Therefore
b−a
π
= ∆x =
4n
n
π
yields b − a = π4 . So, the length of the interval
[ π ] of integration equals 4 . We
suspect then that the integral of integration is 0, 4 , but we are not sure yet of this
assumption. ∑
( iπ )
Now, in the -notation we have the general term of the form tan 4n
. Since in
a Riemann sum as the one above the general term is f (xi ), we note that
( )
iπ
f (xi ) = tan
4n
iπ
yields f (x) = tan(x), and xi = 4n
.
The actual endpoints of integration (a and b) are determined by noting that
a = x0 and b = xn . Therefore,
π
a = x0 = 0, while b = xn = .
4
This concludes that indeed
( ) ∫ π
n
4
π ∑
iπ
tan(x) dx .
lim
·
tan
=
n→∞ 4n
4n
0
i=1
Problem 2. As in the previous P roblem, we have
f (xi ) =
which yields that f (x) =
1
1+x2 ,
1+
while xi =
1
( i )2 ,
i
n.
n
So, the endpoints of integration are
a = x0 = 0 and b = xn = 1.
Therefore
1∑
1
lim
( i )2 =
n→∞ n
i=1 1 +
n
n
∫
0
1
1
.
1 + x2
Problem 3. The distance is found as the area under the graph of the velocity
function between t = 3 to t = 5. Also note that on this time interval the velocity
is always positive, and thus its graph lies above the x-axis.
So, we split the interval [3, 5] into n equal subintervals. Then the common length
of these subintervals is
2
5−3
= .
∆x =
n
n
PRACTICE PROBLEMS: SET 1
3
Then the division points will be
4
6
2n
2
= 5.
x0 = 3; x1 = 3 + ; x2 = 3 + ; x3 = 3 + · · · xn = 3 +
n
n
n
n
So, the right sum approximating the area under the graph of f (t) between t = 3
and t = 5 is
n
2 ∑
Rn = ·
f (xi )
n i=1
(
)
n
2 ∑
2i
= ·
f 3+
n i=1
n
(
)2
n
2 ∑
2i
= ·
3+
n i=1
n
n
2 ∑
12i 4i2
·
9+
+ 2
n i=1
n
n
(
)
n
n
2
12 ∑
4 ∑ 2
= · 9n +
·
i+ 2 ·
i
n
n i=1
n i=1
(
)
12 n(n + 1)
4 n(n + 1)(2n + 1)
2
·
+ 2·
= · 9n +
n
n
2
n
6
(
)
2
2(n + 1)(2n + 1)
= · 9n + 6(n + 1) +
n
3n
12(n + 1) 4(n + 1)(2n + 1)
= 18 +
+
.
n
3n2
We know that
A = lim Rn
=
n→∞
and so,
A
)
12(n + 1) 4(n + 1)(2n + 1)
+
= lim 18 +
n→∞
n
3n2
(
(
)
(
)(
))
1
4
1
1
= lim 18 + 12 1 +
+
1+
2+
n→∞
n
3
n
n
4
= 18 + 12 + · 2
3
98
=
3
(
Problem 4. We split the interval [−2, 4] into n equal subintervals, each of length
= n6 . The division points are
4−(−2)
n
6
12
< x2 = −2 +
< · · · < xn = 4,
n
n
where for each i, we have xi = −2 + 6i
n . Therefore
((
)
)2
∫ 4
n
6∑
6i
2
(x − 5) dx = lim
−2 +
−5 .
n→∞ n
n
−2
i=1
x0 = −2 < x1 = −2 +
4
MATH 101: PROF. DRAGOS GHIOCA
We evaluate
(
)2
6i
24i 36i2
=4−
+ 2 ,
n
n
n
and therefore the above sum equals
((
)
)2
n
∑
6i
−2 +
−5
n
i=1
)
n (
∑
24i 36i2
=
−1 −
+ 2
n
n
i=1
−2 +
= −
n
∑
1−
i=1
n
n
24 ∑
36 ∑ 2
i+ 2
i .
n i=1
n i=1
We evaluate each of the above three sums:
n
∑
1 = n,
i=1
n
∑
i=1
n
∑
i2 =
i=1
i=
n(n + 1)
,
2
n(n + 1)(2n + 1)
.
6
So, using these computations in the above sum we obtain that
((
)
)2
n
∑
6i
−2 +
−5
n
i=1
24n(n + 1) 36n(n + 1)(2n + 1)
+
2n
6n2
6(n + 1)(2n + 1)
−n − 12(n + 1) +
n
12n2 + 18n + 6
−n − 12n − 12 +
n
6
−13n − 12 + 12n + 18 +
n
6
−n + 6 + .
n
= −n −
=
=
=
=
Therefore
∫
4
−2
(x2 − 5) dx
=
=
=
=
((
n
6∑
lim
−2 +
n→∞ n
i=1
(
6
· −n + 6 +
lim
n→∞ n
36 36
lim −6 +
+ 2
n→∞
n
n
−6.
)
)2
6i
−5
n
)
6
n