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Material Type: Exam; Professor: Wills; Class: SI Calculus I; Subject: Math; University: Weber State University; Term: Unknown 2006;
Typology: Exams
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Mathematics 1210 Autumn 2006 Instructor: Mike Wills
Practice Midterm 3 Solutions Problem #1 (5 points):
Let f (x) = 2 cos x sin x. Find the indefinite integral of f. Solution By the half angle formula,
(1)
2 cos x sin xdx =
sin(2x)dx = −
cos(2x) 2
where C is an arbitrary constant.
Problem #2 (5 points):
It costs Cut-Me-Own-Throat Dibbler 4 + 3x + x^2 Ankh-Morpork dollars to procure 100x sausage-inna-buns. Find Dibbler’s average cost and marginal cost. What procurement level will minimize the average cost?
Let C(x) = 4 + 3x + x^2. Then the average cost is given by A(x) = C(x) x =^
4 x + 3 +^ x. The marginal cost will be given by^ c(x) =^ C
′(x) =
3 + 2x. The minimum cost occurs when
c(x) = A(x)
3 + 2x =
x
x =
x x^2 = 4.
Since x ≥ 0 we conclude that x = 2. Thus, when Dibbler procures 200 sausage-inna-buns, his cost is minimized.
Problem #3 (5 points): Carefully define what it means for the function f to be integrable on the interval [a, b].
Solution Let n ∈ N. Let ∆x = b−n a. For i = 0, 1 ,... , n, define xi = a + i∆x.
For i = 1,... , n, let x∗ i ∈ [xi− 1 , xi]. Let Sn =
∑^ n
i=
f (x∗ i )∆x. We say
that f is integrable if lim n→∞
Sn exists.
Problem #4 (5 points):
Compute
1 (x
(^3) − √x)dx.
Solution We compute: ∫ (^4)
1
(x^3 − x
1 (^2) )dx = x
4 4
2 x
(^32)
3
4 1
=