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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.R.105d

Area functions and the Fundamental Theorem Consider the function
ƒ(t) = { t      if  ―2 ≤ t < 0
t²/2    if    0 ≤ t ≤ 2
and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(d) Evaluate F ' (―1) and F ' (1). Interpret these values.

Guida verificata passo dopo passo
1
Step 1: Recall the Fundamental Theorem of Calculus, which states that if F(x) = ∫ₐˣ ƒ(t) dt, then F'(x) = ƒ(x). This means the derivative of the area function F(x) is equal to the value of the function ƒ(x) at x.
Step 2: To evaluate F'(−1), observe that F'(x) = ƒ(x). From the graph and the piecewise definition of ƒ(t), for −2 ≤ t < 0, ƒ(t) = t. Therefore, ƒ(−1) = −1.
Step 3: To evaluate F'(1), observe again that F'(x) = ƒ(x). From the graph and the piecewise definition of ƒ(t), for 0 ≤ t ≤ 2, ƒ(t) = t²/2. Therefore, ƒ(1) = (1²)/2 = 1/2.
Step 4: Interpret the values: F'(−1) = −1 indicates that at x = −1, the rate of change of the area function F(x) is equal to the value of ƒ(t) at t = −1, which is −1. Similarly, F'(1) = 1/2 indicates that at x = 1, the rate of change of the area function F(x) is equal to the value of ƒ(t) at t = 1, which is 1/2.
Step 5: The values of F'(−1) and F'(1) provide insight into how the function ƒ(t) contributes to the accumulation of area in F(x) at specific points. Negative values of ƒ(t) (e.g., at t = −1) reduce the accumulated area, while positive values (e.g., at t = 1) increase it.

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