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In the example shown above, why do we see the notation "f (x) · f (x)" as part of Step 1? f (x) · f (x)= u · v. In our problem, u = f (x), and dv = f' (x) dx, so v = f (x). This is the antiderivative of the given integrand, f (x) · f (x)=S ƒ (x) · f' (æ) dæ . • f (x) · f (x)=v du. In our problem, u = f (x), and dv = f' (x) dx.

In the example shown above, why do we see the notation "f (x) · f (x)" as part of Step 1? f (x) · f (x)= u · v. In our problem, u = f (x), and dv = f' (x) dx, so v = f (x). This is the antiderivative of the given integrand, f (x) · f (x)=S ƒ (x) · f' (æ) dæ . • f (x) · f (x)=v du. In our problem, u = f (x), and dv = f' (x) dx.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
ChapterA: Appendix
SectionA.2: Geometric Constructions
Problem 10P: A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in...
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Let f and f' be continuous functions on [a, b]. Use integration by parts to show that f. f (x) f' (¤) dæ = } (f(b)² – f(a)²). In the steps that follow, we will use u = f (x), dv= f' (x) dæ. The next question asks about these steps. Step 1: Apply u dv = uv, – L v du Step 1: S. f (x) f' (x) dx = f (x) · f (x) . – S. f (2) f' (2) dæ Step 2: Gather like terms. Step 2: 2 S. f (x) f' (2) dr = f (x) · f (x) 2 Step 3: Plug in the limits of integration, and subtract. Step 3: 2 f f (x) f' (z) da = f(b)² – f(a)² Step 4: Divide both sides of the equation by 2. Step 4: S. f (2) f' (2) dz =}(5(b)² – f(a)°) (50)² – f(@})
Transcribed Image Text:Let f and f' be continuous functions on [a, b]. Use integration by parts to show that f. f (x) f' (¤) dæ = } (f(b)² – f(a)²). In the steps that follow, we will use u = f (x), dv= f' (x) dæ. The next question asks about these steps. Step 1: Apply u dv = uv, – L v du Step 1: S. f (x) f' (x) dx = f (x) · f (x) . – S. f (2) f' (2) dæ Step 2: Gather like terms. Step 2: 2 S. f (x) f' (2) dr = f (x) · f (x) 2 Step 3: Plug in the limits of integration, and subtract. Step 3: 2 f f (x) f' (z) da = f(b)² – f(a)² Step 4: Divide both sides of the equation by 2. Step 4: S. f (2) f' (2) dz =}(5(b)² – f(a)°) (50)² – f(@})
In the example shown above, why do we see the notation "f (x) · ƒ (x)" as part of Step 1? f (x) · f (x)= u · v. In our problem, u = f (x), and dv = f' (x) dx, so v = f (x). This is the antiderivative of the given integrand, f (x) · f (x)=S ƒ (x) · f' (x) dx . f (x) · ƒ (x)=v du. In our problem, u = f (x), and dv = f' (x) dx. So v = f (x) and du = f' (x).
Transcribed Image Text:In the example shown above, why do we see the notation "f (x) · ƒ (x)" as part of Step 1? f (x) · f (x)= u · v. In our problem, u = f (x), and dv = f' (x) dx, so v = f (x). This is the antiderivative of the given integrand, f (x) · f (x)=S ƒ (x) · f' (x) dx . f (x) · ƒ (x)=v du. In our problem, u = f (x), and dv = f' (x) dx. So v = f (x) and du = f' (x).
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