Due: 03/31/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #10
Name:
Question:
1
2
3
4
5
Total
Points:
8
8
14
20
10
60
Score:
Instructions: Please show all of your work as partial credit will be given where appropriate, and there may
be no credit given for problems where there is no work shown. All answers should be completely simplified,
unless otherwise stated. No calculators or electronics of any kind are allowed.
1. (8 points) (Area of left polygons) Sketch the graph of f (x) = −x + 5 over the interval [1, 5]; then divide
[1, 5] into 4 equal subintervals and sketch the polygons using the left endpoint, that . Finally, calculate
the area of the corresponding polygon.
Solution:
8
7
6
5
4
3
2
1
0
-1 0
3
X
1
f (1 + i) =
i=0
2
3
X
3
4
(−i + 4) = −
i=0
5
6
3·4
+ 16 = 10
2
2. (8 points) (Area of right polygons) Sketch the graph of f (x) = x2 − 4x + 5 over the interval [0, 4]; then
divide [0, 4] into 4 equal subintervals and sketch the polygons using the right endpoint. Finally, calculate
the area of the corresponding inscribed polygon.
Solution:
8
7
6
5
4
3
2
1
0
-1 0
1
2
3
1
4
5
6
MATH 1210, Spring 2016
Lab #10
4
X
f (i) =
3
X
i=1
(i2 − 4i + 5) =
i=0
3·4
3·4·7
−4
+ 20 = 14 − 24 + 20 = 10
6
2
3. (Riemann sum) Find the area of the region under the curve y = f (x) over the interval [a, b]. To do
this, divide the interval [a, b] into n equal subintervals, calculate the area of the corresponding polygon
using the left endpoint,and then let n → ∞.
(a) (7 points) y = x + 2 ; a = 1, b = 4
Solution:
∆x =
n−1
X
f (xi )∆x =
i=0
3
n
xi = 1 + i∆x
n−1
X
(3 + i∆x)∆x = 3n∆x + ∆x2
i=0
n(n − 1)
9(n − 1) n→∞
9
27
=9+
−→ 9 + =
2
2n
2
2
(b) (7 points) f (x) = 2x2 + 1 ; a = 0, b = 4
Solution:
∆x =
n−1
X
f (xi )∆x =
4
n
n−1
X
(2(i∆x)2 + 1)∆x = 2(∆x)3
i=0
i=0
=
xi = i∆x
n(n − 1)(2(n − 1) + 1)
+ n∆x
6
128n(n − 1)(2n − 1)
128 + 12
140
n→∞ 128 · 2
+ 4 −→
+4=
=
3
6n
6
3
3
4. (Infinite sums) Associate the limit of the sums to a definite integral and solve it using the 2nd
fundamental theorem of calculus.
n
X
5i 5
(a) (5 points) lim
( )5
n→∞
n n
i=1
Solution: Let xi =
5i
n,
then ∆x =
5
n
and x0 = 0 and xn = 5. So
Z 5
n
n
X
X
5i 5 5
5
lim
( )
= lim
(xi ) ∆x =
x5 dx
n→∞
n→∞
n
n
0
i=1
i=1
(b) (5 points) lim
n→∞
n
X
(
i=−n+1
Solution: Let xi =
2i 3 2
)
n n
2i
n,
then ∆x =
2
n
and x0 = 0 and xn = 2. So
n
X
Z 2
n
X
2i 3 2
3
lim
( )
= lim
(xi ) ∆x =
x3 dx
n→∞
n→∞
n
n
0
i=−n+1
i=1
2
Due: 03/31/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #10
Name:
n
X
4i
2i
i
2
(c) (5 points) lim
(1 −
+ ( )2 − ( )3 )
n→∞
n
n
n
n
i=1
Solution: Let xi =
i
n,
then ∆x =
1
n
and x0 = 0 and xn = 1. So
n
n
X
X
4i
2i
i
2
(1 −
+ ( )2 − ( )3 ) = lim
(1 − 4xi + 4x2i x3i )2∆x
n→∞
n→∞
n
n
n
n
i=1
i=1
lim
Z
1
=
2(1 − 4x + 4x2 − x3 )dx
0
(d) (5 points) lim
n
X
n→∞
√
i=1
√
√
2πi
2πi 2 2π
cos((
) )
n
n
n
√
Solution: Let xi =
lim
n→∞
n
X
i=1
√
2πi
n ,
√
then ∆x =
2π
n
and x0 = 0 and xn =
√
2π. So
√
√
Z √2π
n
X
2πi
2πi 2 2π
2
cos((
) )
= lim
xi cos(xi )∆x =
x cos(x2 )dx
n→∞
n
n
n
0
i=1
5. (Linearity of the integral)
(a) (5 points) Knowing that
R2
0
2f (x) + 3dx = 8,
R2
0
f (x) − g(x)dx = −1, calculate
R2
0
g(x)dx.
R2
R2
R2
Solution: We can rewrite the fist integral as 2 0 f (x)dx+ 0 3dx = 8, thus we get 0 f (x)dx =
R2
R2
R2
1. We can rewrite the second integral as 0 f (x) − 2 g(x)dx = −1, which gives 0 g(x) = 2.
(b) (5 points) We know that that
Determine f (1).
R1
0
3x2 − f (x)dx = 0, f (x) is constant for x ≥ 1 and
R2
0
f (x)dx = 0.
R1
R1
Solution: x3 is an antiderivative of 3x2 , so 0 3x2 dx = [x3 ]10 = 1. Thus 0 f (x)dx = 1. If we
R1
R2
split the second integral given and substitute 1 for 0 f (x)dx, we get 1 + 1 f (x)dx = 0. Hence
R2
f (x)dx = −1. But it has to be the integral of a constant, say c, so we get (2 − 1)c = −1. It
1
gives us c = −1 and thus f (1) = −1.
3