Area functions Let A(x) be the area of the region bounded by the t -axis and the graph of y=ƒ(t) from t=0 to t=x. Consider the following functions and graphs.

a. Find A(2) .

ƒ(t) = {-2t+8 if t ≤ 3 ; 2 if t >3 <IMAGE>

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Identify the function \( f(t) \) and the intervals: \( f(t) = -2t + 8 \) for \( t \leq 3 \) and \( f(t) = 2 \) for \( t > 3 \).

Since we need to find \( A(2) \), we focus on the interval from \( t = 0 \) to \( t = 2 \), where \( f(t) = -2t + 8 \).

Set up the integral to find the area under the curve from \( t = 0 \) to \( t = 2 \): \( A(2) = \int_{0}^{2} (-2t + 8) \, dt \).

Calculate the integral: Find the antiderivative of \( -2t + 8 \), which is \( -t^2 + 8t \).

Evaluate the definite integral by substituting the limits: \( A(2) = [-t^2 + 8t]_{0}^{2} \).

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주요 개념

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Definite Integral

The area function A(x) represents the definite integral of the function f(t) from t=0 to t=x. This integral calculates the net area between the curve and the t-axis, accounting for regions above and below the axis. Understanding how to evaluate definite integrals is crucial for finding specific area values, such as A(2) in this case.

Piecewise Functions

The function f(t) is defined piecewise, meaning it has different expressions based on the value of t. For t ≤ 3, f(t) = -2t + 8, and for t > 3, f(t) = 2. Recognizing how to work with piecewise functions is essential for correctly evaluating the integral over the specified interval, as the function's behavior changes at t=3.

Area Under a Curve

The area under a curve can be interpreted as the integral of the function over a given interval. In this problem, calculating A(2) involves finding the area under f(t) from t=0 to t=2. This requires integrating the appropriate expression of f(t) over the specified limits, which is fundamental in applications of calculus to determine physical quantities like area.

Area functions Let A(x) be the area of the region bounded by the t -axis and the graph of y=ƒ(t) from t=0 to t=x. Consider the following functions and graphs.a. Find A(2) .ƒ(t) = {-2t+8 if t ≤ 3 ; 2 if t >3 <IMAGE>