IV. Problem Solving: A. Write the expression as a single Integral = xp(x)f S +xp(x)fT 2. f f(x)dx – , f(x)dx = 3. f f(x)dx + f, f(x)dx = 4. L, f(x)dx + f, f(x)dx = 5. (x)dx – f° f(x)dx = B. Assume that Si f(x)dx = 3 and ig(x)dx = -2, find the following: %3D 1. Sf(x) + g(x)]dx = 2. 12f(x) – g(x)] dx = 3. Sf(x) – g(x)]dx = %3D 4. 14g(x) – 3f(x)] dx = 5. f(x)+3g(x)] dx =

Calculus: Early Transcendentals
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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IV. Problem Solving:
A. Write the expression as a single Integral
1. f f(x)dx+ [} f(x)dx =,
2. f f(x)dx – ſ, f(x)dx =
3. f f(x)dx + f, f(x)dx =
4. L, f(x)dx + f, f(x)dx =
5. (x)dx – £° f(x)dx =
B. Assume that Si f(x)dx = 3 and Sig(x)dx = -2, find the following:
1. f(x) + g(x)]dx =
2. 12f(x) – g(x)] dx =
3. Sf(x) – g(x)]dx =
%3D
4. 14g(x) – 3f(x)] dx =
5. f(x)+3g(x)] dx =
Transcribed Image Text:IV. Problem Solving: A. Write the expression as a single Integral 1. f f(x)dx+ [} f(x)dx =, 2. f f(x)dx – ſ, f(x)dx = 3. f f(x)dx + f, f(x)dx = 4. L, f(x)dx + f, f(x)dx = 5. (x)dx – £° f(x)dx = B. Assume that Si f(x)dx = 3 and Sig(x)dx = -2, find the following: 1. f(x) + g(x)]dx = 2. 12f(x) – g(x)] dx = 3. Sf(x) – g(x)]dx = %3D 4. 14g(x) – 3f(x)] dx = 5. f(x)+3g(x)] dx =
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