Textbook Question
Area functions The graph of Ζ is shown in the figure. Let A(x) = β«βΛ£ Ζ(t) dt and F(x) = β«βΛ£ Ζ(t) dt be two area functions for Ζ. Evaluate the following area functions.
(a) A(2)
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9. Graphical Applications of Integrals
Area Between Curves
2:56 minutesProblem 5.3.17b
Textbook Question
Area functions for the same linear function Let Ζ(t) = t and consider the two area functions A(π) = β«βΛ£ Ζ(t) dt and F(π) = β«βΛ£ Ζ(t) dt .
(b) Evaluate F(4) and F(6). Then use geometry to find an expression for F (π) , for π β₯ 2.
Verified step by step guidance1
Step 1: Understand the problem. We are given a linear function Ζ(t) = t and two area functions: A(π) = β«βΛ£ Ζ(t) dt and F(π) = β«βΛ£ Ζ(t) dt. The goal is to evaluate F(4) and F(6), and then derive a general expression for F(π) using geometry for π β₯ 2.
Step 2: Recall the geometric interpretation of definite integrals. The integral β«βΛ£ Ζ(t) dt represents the area under the curve Ζ(t) = t from t = 2 to t = π. Since Ζ(t) = t is a linear function, the graph is a straight line, and the area can be calculated as the area of a trapezoid or triangle.
Step 3: Evaluate F(4). To find F(4), compute the definite integral β«ββ΄ t dt. This represents the area under the line Ζ(t) = t from t = 2 to t = 4. Use the formula for the definite integral of t, β« t dt = (1/2)tΒ², and evaluate it at the bounds 2 and 4.
Step 4: Evaluate F(6). Similarly, compute the definite integral β«ββΆ t dt. This represents the area under the line Ζ(t) = t from t = 2 to t = 6. Again, use the formula for the definite integral of t, β« t dt = (1/2)tΒ², and evaluate it at the bounds 2 and 6.
Step 5: Derive a general expression for F(π) for π β₯ 2. Using geometry, note that the area under the line Ζ(t) = t from t = 2 to t = π forms a trapezoid or triangle. The formula for the area of a trapezoid is (1/2)(base1 + base2) Γ height. Alternatively, use the definite integral formula β«βΛ£ t dt = (1/2)πΒ² - (1/2)(2Β²) to express F(π).
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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 defined by a function over a specific interval. In this context, the area functions A(x) and F(x) are calculated using the definite integral of the function f(t) = t from a lower limit to an upper limit. Understanding how to evaluate definite integrals is crucial for finding the values of F(4) and F(6).
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Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of differentiation with integration, stating that if F is an antiderivative of f on an interval, then the integral of f from a to b is equal to F(b) - F(a). This theorem is essential for evaluating the area functions A(x) and F(x) and helps in deriving expressions for these functions based on their limits.
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Geometric Interpretation of Integrals
The geometric interpretation of integrals involves visualizing the area under a curve as a physical space that can be calculated. In this problem, using geometry to find an expression for F(x) for x β₯ 2 means recognizing the shape formed by the linear function f(t) = t and calculating the area of the resulting geometric figure, such as a triangle or trapezoid, to derive a formula for F(x).
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