HOMEWORK 12 FOR MATH 2250 - MODEL SOLUTION
Every problem will be worth two points.
(1) Use finite approximation to estimate the area under the graph of f (x) = 4 − x2
between x = −2 and x = 2.
(a) A lower sum with two rectangles of equal width.
Approximation: 0 · 2 + 0 · 2 = 0.
(b) A lower sum with four rectangles of equal width.
Approximation: 0 · 1 + 3 · 1 + 3 · 1 + 0 · 1 = 6.
Date: April 21, 2012.
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2
HOMEWORK 12 FOR MATH 2250 - MODEL SOLUTION
(c) An upper sum with two rectangles of equal width.
Approximation: 4 · 2 + 4 · 2 = 16.
(d) An upper sum with four rectangles of equal width.
Approximation: 3 · 1 + 4 · 1 + 4 · 1 + 3 · 1 = 14.
(2) Evaluate the following sums.
36
X
(a)
k.
k=9
36
X
k=9
k=
36
X
k=1
k−
8
X
k=1
k=
36(36 + 1) 8(8 + 1)
−
= 666 − 36 = 630.
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HOMEWORK 12 FOR MATH 2250 - MODEL SOLUTION
(b)
17
X
3
k2.
k=3
17
X
k
2
=
k=3
17
X
2
k −
k=1
2
X
k2 =
k=1
17(17 + 1)(2 · 17 + 1) 2(2 + 1)(2 · 2 + 1)
−
6
6
= 1785 − 5 = 1780.
(c)
71
X
k(k − 1).
k=18
71
X
k(k − 1) =
k=18
71
X
2
(k − k) =
k=18
=
71
X
2
k −
k=18
71
X
k2 −
17
X
71
X
−
k=1
k=1
k
k=18
!
k2
71
X
k=1
k−
17
X
!
k
k=1
71(71 + 1)(2 · 71 + 1) 17(17 + 1)(2 · 17 + 1) 71(71 + 1) 17(17 + 1)
−
−
+
6
6
2
2
= 121836 − 1785 − 2556 + 153 = 117648.
=
(3) Graph the integrands and use areas to evaluate the integral.
Z
1
(1 − |x|)dx.
−1
Z
1
(1 − |x|)dx = Area of =
−1
1
· 2 · 1 = 1.
2
(4) Graph f (t) = t2 − t and find its average value over [−2, 1].
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HOMEWORK 12 FOR MATH 2250 - MODEL SOLUTION
Z
1
Z
2
Z
1
2
Z
1
t − tdt =
t dt −
tdt
−2
−2
−2
3
2
1
(−2)3
(−2)2
1
9
=
−
−
−
= .
3
3
2
2
2
Z 1
1
1 9
3
av(f ) =
f (t)dt = · = .
1 − (−2) −2
3 2
2
f (t)dt =
−2
1