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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.3.107b
Capitolo 5, Problema 5.3.107b

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.                                                                          
                                                                                                                                                                                     (b) Suppose ƒ is a negative increasing function, for 𝓍 > 0 . Then the area function A(𝓍) = ∫₀ˣ ƒ(t) dt is a decreasing function of 𝓍 .

Guida verificata passo dopo passo
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Step 1: Begin by understanding the problem statement. The function ƒ(t) is given as a negative and increasing function for x > 0. This means that ƒ(t) < 0 and its derivative ƒ'(t) > 0 (since it is increasing). The area function A(x) is defined as A(x) = ∫₀ˣ ƒ(t) dt, which represents the accumulated area under the curve of ƒ(t) from t = 0 to t = x.
Step 2: To determine whether A(x) is a decreasing function, compute the derivative of A(x) with respect to x. Using the Fundamental Theorem of Calculus, we know that A'(x) = ƒ(x). This derivative tells us the rate of change of the area function A(x) with respect to x.
Step 3: Analyze the sign of A'(x). Since ƒ(x) is negative for x > 0 (as given in the problem), A'(x) = ƒ(x) is also negative. A negative derivative implies that the function A(x) is decreasing for x > 0.
Step 4: Consider the behavior of ƒ(t) being an increasing function. Although ƒ(t) is increasing, it remains negative for all x > 0. This means that while the rate at which A(x) decreases might slow down (due to ƒ(t) becoming less negative), A(x) will still continue to decrease because ƒ(x) < 0.
Step 5: Conclude that the statement is true. The area function A(x) = ∫₀ˣ ƒ(t) dt is indeed a decreasing function of x for x > 0, because its derivative A'(x) = ƒ(x) is negative for all x > 0.

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Definite Integral

A definite integral represents the signed area under a curve defined by a function over a specific interval. It is calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. The value of a definite integral can provide insights into the accumulation of quantities, such as area, volume, or total change, over the interval.
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Definition of the Definite Integral

Increasing and Decreasing Functions

A function is considered increasing on an interval if, for any two points within that interval, the function's value at the second point is greater than or equal to its value at the first point. Conversely, a function is decreasing if its value diminishes as the input increases. Understanding these properties is crucial for analyzing the behavior of functions and their integrals.
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Area Function

The area function A(x) = ∫₀ˣ f(t) dt represents the accumulated area under the curve of the function f(t) from 0 to x. The behavior of this function can be influenced by the properties of f(t), such as whether it is positive or negative. In the case of a negative increasing function, the area function will decrease as x increases, reflecting the negative contributions to the area.
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Finding Area When Bounds Are Not Given
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