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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 5.3.14a

Area functions The graph of ƒ is shown in the figure. Let A(x) = ∫₀ˣ ƒ(t) dt and F(x) = ∫₂ˣ ƒ(t) dt be two area functions for ƒ. Evaluate the following area functions.
(a) A(2)

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Step 1: Understand the problem. We are tasked with evaluating A(2), where A(x) = ∫₀ˣ ƒ(t) dt. This represents the area under the curve of ƒ(t) from t = 0 to t = x.
Step 2: Analyze the graph. The graph shows the function ƒ(t) and the areas of specific regions. From t = 0 to t = 1, the area is shaded green and labeled as 'Area = 8'. This area is above the t-axis, so it contributes positively to the integral.
Step 3: From t = 1 to t = 2, the graph dips below the t-axis, and the area is shaded green and labeled as 'Area = 5'. Since this region is below the t-axis, it contributes negatively to the integral.
Step 4: To compute A(2), sum the contributions of the areas from t = 0 to t = 2. Specifically, add the positive area from t = 0 to t = 1 and subtract the negative area from t = 1 to t = 2.
Step 5: Write the expression for A(2): A(2) = (Area from t = 0 to t = 1) - (Area from t = 1 to t = 2). Substitute the values from the graph: A(2) = 8 - 5.

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Key Concepts

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

Definite Integral

A definite integral represents the signed area under a curve between two points on the x-axis. It is calculated using the integral symbol and limits of integration, providing a numerical value that corresponds to the net area, accounting for areas above and below the x-axis.
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Definition of the Definite Integral

Area Function

An area function, such as A(x) = ∫₀ˣ ƒ(t) dt, defines the accumulated area under the curve of a function ƒ from a starting point to a variable endpoint x. This function helps in understanding how the area changes as x varies, and it is crucial for evaluating specific area values at given points.
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Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, stating that if F is an antiderivative of f on an interval, then the definite integral of f from a to b can be computed as F(b) - F(a). This theorem is essential for evaluating area functions and understanding their properties.
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Related Practice
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(a) Describe the motion of the object over the interval [0,6].

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Use Table 5.6 to evaluate the following indefinite integrals.                                                                                                               

                                                                                                                                                                  

 (a) ∫ e¹⁰ˣ d𝓍

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(a) Divide the interval [34, 64] into n = 5 subintervals with the gridpoints x₀ = 34 , x₁ = 40 , x₂ = 46 , x₃ = 52 , x₄ = 58 , and x₅ = 64. Use left and right Riemann sums to estimate how far Felix fell while traveling at supersonic speed.

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Textbook Question

Area functions for the same linear function Let ƒ(t) = 2t ― 2 and consider the two area functions A (𝓍) = ∫₁ˣ ƒ(t) dt and F(𝓍) = ∫₄ˣ ƒ(t) dt .

(a) Evaluate A (2) and A (3). Then use geometry to find an expression for A (𝓍) , for 𝓍 ≥ 1 .

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(a) ∫¹₋₁ 𝓍ƒ(𝓍²) d𝓍

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Textbook Question

Planetary orbits The planets orbit the Sun in elliptical orbits with the Sun at one focus (see Section 12.4 for more on ellipses). The equation of an ellipse whose dimensions are 2a in the 𝓍-direction and 2b in the y-direction is (𝓍²/a²) + (y² /b²) = 1.

(a) Let d² denote the square of the distance from a planet to the center of the ellipse at (0, 0). Integrate over the interval [ ―a, a] to show that the average value of d² is (a² + 2b²) /3 .

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