Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
8. Definite Integrals
Fundamental Theorem of Calculus
Problem 5.3.11
Textbook Question
Evaluate ∫₃⁸ ƒ ′(t) dt , where ƒ ′ is continuous on [3, 8], ƒ(3) = 4, and ƒ(8) = 20 .

1
Recognize that the integral ∫₃⁸ ƒ ′(t) dt represents the net change of the function ƒ(t) over the interval [3, 8]. This is based on the Fundamental Theorem of Calculus, which states that ∫ₐᵇ ƒ ′(x) dx = ƒ(b) - ƒ(a).
Identify the given values: ƒ(3) = 4 and ƒ(8) = 20. These represent the values of the function ƒ(t) at the endpoints of the interval [3, 8].
Apply the Fundamental Theorem of Calculus: Substitute the values of ƒ(8) and ƒ(3) into the formula ƒ(b) - ƒ(a). Specifically, calculate ƒ(8) - ƒ(3).
Set up the subtraction: ƒ(8) - ƒ(3) = 20 - 4. This represents the net change of the function ƒ(t) over the interval [3, 8].
Conclude that the integral ∫₃⁸ ƒ ′(t) dt is equal to the result of the subtraction performed in the previous step, which represents the total change in ƒ(t) over the interval.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration, stating that if a function is continuous on an interval, the integral of its derivative over that interval equals the difference in the function's values at the endpoints. Specifically, ∫ₐᵇ f'(t) dt = f(b) - f(a). This theorem is essential for evaluating definite integrals involving derivatives.
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Definite Integral
A definite integral represents the signed area under a curve defined by a function over a specific interval [a, b]. It is calculated as the limit of Riemann sums and provides a numerical value that reflects the accumulation of quantities, such as area or total change, between the two bounds. In this case, it helps determine the total change in the function f from t=3 to t=8.
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Continuous Function
A continuous function is one that does not have any breaks, jumps, or holes in its graph over a given interval. For the Fundamental Theorem of Calculus to apply, the derivative f'(t) must be continuous on the interval [3, 8]. This ensures that the integral can be evaluated reliably and that the function f is well-defined at the endpoints.
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Intro to Continuity
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