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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.R.105b

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.
fig
(b) Use the Fundamental Theorem to find an expression for F '(𝓍) for ―2 ≤ 𝓍 < 0.

Verified step by step guidance
1
The problem asks us to find F'(𝓍) for the interval -2 ≤ 𝓍 < 0 using the Fundamental Theorem of Calculus. Recall that the Fundamental Theorem states that if F(𝓍) = ∫ₐˣ ƒ(t) dt, then F'(𝓍) = ƒ(𝓍), provided ƒ is continuous at 𝓍.
The given function ƒ(t) is piecewise defined as ƒ(t) = t for -2 ≤ t < 0 and ƒ(t) = t²/2 for 0 ≤ t ≤ 2. For the interval -2 ≤ 𝓍 < 0, we are concerned only with the first piece of the function, ƒ(t) = t.
Using the Fundamental Theorem, F'(𝓍) = ƒ(𝓍). Since ƒ(t) = t in this interval, we substitute 𝓍 for t, giving F'(𝓍) = 𝓍 for -2 ≤ 𝓍 < 0.
Verify that the function ƒ(t) = t is continuous on the interval -2 ≤ t < 0. This is true because the function t is a simple linear function, which is continuous everywhere.
Thus, the expression for F'(𝓍) in the interval -2 ≤ 𝓍 < 0 is F'(𝓍) = 𝓍.

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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 a function f on an interval [a, b], then the integral of f from a to b can be computed as F(b) - F(a). This theorem also implies that the derivative of the integral function F(x) is equal to the original function f evaluated at x, which is crucial for finding expressions like F'(x).
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Definite Integral

A definite integral represents the signed area under the curve of a function f(t) from a lower limit a to an upper limit b. It is denoted as ∫_a^b f(t) dt and provides a numerical value that corresponds to the accumulation of quantities, such as area, over the specified interval. In this context, F(x) and G(x) are defined as definite integrals of the function f(t) over different intervals.
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Piecewise Function

A piecewise function is defined by different expressions based on the input value. In this case, the function f(t) is defined differently for the intervals t < 0 and 0 ≤ t ≤ 2. Understanding how to evaluate piecewise functions is essential for correctly applying the Fundamental Theorem of Calculus, especially when determining derivatives or integrals over specific intervals.
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Related Practice
Textbook Question

Limit definition of the definite integral Use the limit definition of the definite integral with right Riemann sums and a regular partition to evaluate the following definite integrals. Use the Fundamental Theorem of Calculus to check your answer. 


∫₀² (𝓍²―4) d𝓍

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Textbook Question

Use geometry and properties of integrals to evaluate the following definite integrals.

∫₄⁰ (2𝓍 + √(16―𝓍²)) d𝓍 . (Hint: Write the integral as sum of two integrals.)

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Textbook Question

Integration by Riemann sums Consider the integral ∫₁⁴ (3𝓍― 2) d𝓍.


(c) Evaluate the definite integral by taking the limit as n →∞ of the Riemann sum in part (b).

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.

(f) ∫ₐᵇ (2 ƒ(𝓍) ―3g (𝓍)) d𝓍 = 2 ∫ₐᵇ ƒ(𝓍) d𝓍 + 3 ∫₆ᵃ g(𝓍) d𝓍 .

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.

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Textbook Question

Evaluating integrals Evaluate the following integrals.                                                                                                                                         

                                                                                                                                                                    

 ∫ (cos 7ω) /(16 + sin² 7ω) dω 

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