AP Calculus AB: 9.4.4 The Fundamental Theorem of Calculus, Part II
This section explains Part II of the Fundamental Theorem of Calculus, which provides a powerful way to evaluate definite integrals by using antiderivatives. It connects the calculation of areas under curves with evaluating integrals, showing how the difference of antiderivative values at the interval endpoints gives the exact area, even for complex regions.
The Fundamental Theorem of Calculus, Part II
Let f be defined on the interval [a, b]. The definite integral of f from a to b is if exists.
The fundamental theorem of calculus links the velocity and area problems. It enables you to evaluate definite integrals, thereby finding the area between a curve and the x-axis.
The fundamental theorem of calculus states that if f is continuous on [a, b] and F is an antiderivative of f on that interval, then
Key Terms
The Fundamental Theorem of Calculus, Part II
Let f be defined on the interval [a, b]. The definite integral of f from a to b is if exists.
The fundamental theorem of cal...
note
The fundamental theorem of calculus provides a means of evaluating definite integrals. These are the integrals that are associated with cal...
What is the area between the curve y = x^ 3 + x and the x‑axis on the interval [0, 1]?
A=3/4
What is the area of the region bound between the curve y = x ^2, the line x = 3, and the x‑axis?
A = 9
What is the area between the curve y = x^ 2 + 3 and the x‑axis on the interval [1, 3]?
A=14 2/3
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| Term | Definition |
|---|---|
The Fundamental Theorem of Calculus, Part II |
|
note |
|
What is the area between the curve y = x^ 3 + x and the x‑axis on the interval [0, 1]? | A=3/4 |
What is the area of the region bound between the curve y = x ^2, the line x = 3, and the x‑axis? | A = 9 |
What is the area between the curve y = x^ 2 + 3 and the x‑axis on the interval [1, 3]? | A=14 2/3 |