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Back to FlashcardsMathematics / AP Calculus AB: 9.4.4 The Fundamental Theorem of Calculus, Part II

AP Calculus AB: 9.4.4 The Fundamental Theorem of Calculus, Part II

Mathematics5 CardsCreated 8 months ago

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.

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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

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Key Terms

Term
Definition

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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AP Calculus AB: 9.4.4 The Fundamental Theorem of Calculus, Part II
TermDefinition

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

note

  • The fundamental theorem of calculus provides a means of evaluating definite integrals. These are the integrals that are associated with calculating areas under curves.

  • A simple example involves calculating the area under the line described by f(x) = x on the interval [0, 3]. Since the region involved is a triangle, you can use the area formula for triangles to arrive at the answer.

  • The calculus technique requires you to evaluate the definite integral of f from 0 to 3.

  • Notice that the constant of integration C cancels with itself.

  • The definite integral produces the same result as the area formula.

  • You can use definite integrals to determine areas of more unusual regions. Here you have the area under a parabola.

  • Set up the definite integral of the function f from 0 to 1 and evaluate it.

  • The fundamental theorem tells you to evaluate the
    antiderivative at 1 and subtract the value of the antiderivative at 0.

  • The area of the region is 1/3.

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