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Material Type: Exam; Class: SI Calculus I; Subject: Math; University: Weber State University; Term: Unknown 2008;
Typology: Exams
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Compute the average value of the following functions:
f (x) = x^3 − 5 x^2 + cos(2πx) for 0 ≤ x ≤ 2.
Solution 1. We compute:
fave =
0
f (x)dx
0
(x^3 − 5 x^2 + cos(2πx))dx
[x 4 4
5 x^3 3
sin(2πx) 2 π
f (x) = cos^3 (2x) sin(2x) for π 4 ≤ x ≤ π 2
Solution 2. We compute:
fave =
π
∫ π 2 π 4 f (x)dx
π
∫ π 2 (^4) π^ cos
(^3) (2x) sin(2x)dx
π
[cos (^4) (2x) 8
]π 2 π 4
= −
π
2 π
1
2 MATH 1210 PRACTICE FINAL SOLUTIONS
f (x) = x (^14) − x−^ (^14) for 1 ≤ x ≤ 16.
Solution 3. We compute:
fave =
1
f (x)dx
1
(x (^14) − x−^ (^14) )dx
[ 4 x^54 5
4 x (^34) 3
f (x) = x^19 − 3 x^5 + 43x for − 25 ≤ x ≤ 25.
Solution 4. Notice that f is an odd function and the interval is sym- metric about x = 0. Consequently that average of f on the interval is
