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A (x) = Sí f (t) dt, |3| where f (t) = 4 – 2t. To compute A (x), we apply the first Fundamental Theorem of Calculus: A (x) = Sï 4 – 2t dt = (4t – t²) lï = (4x – a²) – (4 - 1 – 1º). |3D O 6 What is the result? A (x) = 4x – x² – 3 %3D R A (x) = 4x – a2 – 5 R • A (x) = 4x –- a²

A (x) = Sí f (t) dt, |3| where f (t) = 4 – 2t. To compute A (x), we apply the first Fundamental Theorem of Calculus: A (x) = Sï 4 – 2t dt = (4t – t²) lï = (4x – a²) – (4 - 1 – 1º). |3D O 6 What is the result? A (x) = 4x – x² – 3 %3D R A (x) = 4x – a2 – 5 R • A (x) = 4x –- a²

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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we are given the integral function A (x) = Sĩ ƒ (t) dt, %3| where f (t) = 4 – 2t. |3D To compute A (x), we apply the first Fundamental Theorem of Calculus: A (x) = Sĩ 4 – 2t dt = (4t – t²) lỉ = (4x – 2²) – (4 · 1 – 1°). 4·1 %3| %3| What is the result? A (x) = 4x – a² – 3 A (x) = 4x – x² – 5 . A (г) — 4х — а? AUR
Transcribed Image Text:we are given the integral function A (x) = Sĩ ƒ (t) dt, %3| where f (t) = 4 – 2t. |3D To compute A (x), we apply the first Fundamental Theorem of Calculus: A (x) = Sĩ 4 – 2t dt = (4t – t²) lỉ = (4x – 2²) – (4 · 1 – 1°). 4·1 %3| %3| What is the result? A (x) = 4x – a² – 3 A (x) = 4x – x² – 5 . A (г) — 4х — а? AUR
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