Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. (d) If A(π) = 3πΒ²β πβ 3 is an area function for Ζ, then B(π) = 3πΒ² β π is also an area function for Ζ.
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Recall that an area function for a function \( f \) is defined as \( A(x) = \int_a^x f(t) \, dt \) for some fixed lower limit \( a \). The key property is that the derivative of the area function equals the original function, i.e., \( A'(x) = f(x) \).
Given \( A(x) = 3x^2 - x - 3 \), find its derivative to identify \( f(x) \). Using the power rule, \( A'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(x) - \frac{d}{dx}(3) = 6x - 1 - 0 = 6x - 1 \). So, \( f(x) = 6x - 1 \).
Next, check the function \( B(x) = 3x^2 - x \). Find its derivative: \( B'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(x) = 6x - 1 \). Notice that \( B'(x) = f(x) \) as well.
Since both \( A(x) \) and \( B(x) \) have the same derivative \( f(x) = 6x - 1 \), they differ by a constant. To confirm if \( B(x) \) is also an area function for \( f \), check the difference \( A(x) - B(x) = (3x^2 - x - 3) - (3x^2 - x) = -3 \), which is a constant.
Therefore, \( B(x) \) can also serve as an area function for \( f \) because area functions differ by a constant. This aligns with the Fundamental Theorem of Calculus, which states that any two antiderivatives of the same function differ by a constant.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Area Function and the Fundamental Theorem of Calculus
An area function A(x) for a function f is defined as the integral of f from a fixed point to x. According to the Fundamental Theorem of Calculus, the derivative of this area function A'(x) equals the original function f(x). This relationship is key to verifying if a given function is an area function for f.
To check if a function is an area function for f, differentiate it and compare the result to f. Differentiation of polynomials involves applying power rules term-by-term. For example, the derivative of 3xΒ² - x - 3 is 6x - 1, which must match f(x) for the function to be an area function.
Adding or subtracting a constant to an area function does not change its derivative, so it remains an area function for the same f. This means if A(x) is an area function for f, then A(x) + C, where C is any constant, is also an area function for f.