Area functions and the Fundamental Theorem Consider the functionƒ(t) = { t if ―2 ≤ t \u003c 0t²/2 if 0 ≤ t ≤ 2and 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.

Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Ch. 5 - Integration

Problem 5.R.105d

Chapter 5, Problem 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.

Verified step by step guidance

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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then the integral of f from a to b is equal to F(b) - F(a). This theorem allows us to evaluate definite integrals and understand the relationship between a function and its area under the curve.

Derivative of an Integral Function

When evaluating the derivative of an integral function, such as F(x) = ∫₋₁ˣ f(t) dt, we apply the Fundamental Theorem of Calculus. The derivative F'(x) gives us the value of the integrand f evaluated at the upper limit of integration, which provides insight into the rate of change of the area under the curve as x varies.

Piecewise Functions

A piecewise function is defined by different expressions based on the input value. In this case, the function f(t) has two distinct expressions depending on whether t is less than 0 or between 0 and 2. Understanding how to evaluate and differentiate piecewise functions is crucial for correctly applying calculus concepts to such functions.

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Piecewise Functions

Related Practice

Textbook Question

Area versus net area Find (i) the net area and (ii) the area of the region bounded by the graph of ƒ and the 𝓍-axis on the given interval. You may find it useful to sketch the region.

ƒ(𝓍) = 𝓍⁴ ― 𝓍² on [―1, 1]

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Textbook Question

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.

(a) Evaluate F(―2) and F(2).

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Textbook Question

Evaluating integrals Evaluate the following integrals.