Flashcards in 9.4.4 The Fundamental Theorem of Calculus, Part II Deck (5):
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
- 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]?
What is the area of the region bound between the curve y = x ^2, the line x = 3, and the x‑axis?
A = 9