1. Group Work 02/27
(1) Express the integral as a limit of Riemann sums. Do not evaluate the limit.
(a)
3
Z
√
4 + x2 dx
1
Solution:
2
3−1
=
n
n
∆x =
xi = a + i∆x = 1 +
2i
n
Hence
Z
3
√
4+
x2
dx = lim
1
n
X
n→∞
f (xi ) ∆x = lim
n→∞
i=1
n
X
i=1
s
2
2i
2
4+ 1+
·
n
n
(b)
Z
2
5
1
x +
x
2
dx
Solution:
∆x =
5−2
3
=
n
n
xi = a + i∆x = 2 +
3i
n
Hence
Z
2
5
1
x2 +
x
dx = lim
n→∞
n
X
i=1
"
2
n
X
3i
+
f (xi ) ∆x = lim
2+
n→∞
n
i=1
1
1
2+
#
3i
n
·
3
n
(2) Write the integral as a limit of riemann sums and then evaluate the limit
Z 4
(x2 − 4x + 2)dx
1
Solution:
4−1
3
=
n
n
∆x =
xi = a + i∆x = 1 +
3i
n
Hence
4
Z
(x2 − 4x + 2)dx = lim
n→∞
1
"
n
X
i=1
3i
1+
n
2
#
3i
3
−4 1+
+2 ·
n
n
n X
12i
3
6i 9i2
+2 ·
= lim
1+ + 2 − 4+
n→∞
n
n
n
n
i=1
n X
27i2
= lim
n→∞
"
= lim
n→∞
= lim
n→∞
= lim
27
n3
27
6
n→∞
= lim
n→∞
9
2
27
n3
n
X
i=1
i=1
!
2
i
n3
18
− 2
n
n(n + 1)(2n + 1)
6
18i
3
− 2 −
n
n
n
X
!
3
−
n
i
i=1
18
− 2
n
18
−
2
n + 1 2n + 1
·
n
n
1
1
1
1+
2+
− 9 1+
− 3
n
n
n
2
n(n + 1)
2
n
X
1
i=1
n n+1
·
n
n
=
!#
3
− (n)
n
− 3
9
(1)(2) − 9(1) − 3
2
=
−3
(3) Evaluate the integrals by interpreting it in terms of areas
(a)
3
Z
(1 − x) dx
0
Solution:
From [0, 1] we have a triangle of height and base 1 so the area above the x-axis
is A1 = 12 . From [1, 3] we have a triangle of height and base 2 so the area below
the x-axis is A2 = 2. Hence the net area is
Z 3
1
−3
(1 − x) dx = A1 − A2 = − 2 =
2
2
0
(b)
Z
4
|x − 2| dx
−4
Solution:
From [−4, 2] we have a triangle of height and base 6 so the area is 18. From [2, 4]
we have a triangle of height and base 2 so the area is A2 = 2. Since f (x) ≥ 0
then
Z 4
|x − 2| dx = 18 + 2 = 20
−4
(c)
Z
3
√
9 − x2 dx
0
Solution:
x2 + y 2 = 9 is a circle or radius 3 centered at the origin. The integral determines
the top right quarter of the circle and so
Z 3√
9π
9 − x2 dx =
4
0
3
Z
b
f (x)dx :
(4) Write as a single integral in the form
a
Z
2
5
Z
Z
−2
−1
f (x) dx −
f (x) dx +
f (x) dx
−2
2
Solution:
Z
5
f (x)
−1
4