Table of contents

0. Functions7h 52m
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
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
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
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

8. Definite Integrals

Fundamental Theorem of Calculus

Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

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1:58 minutes

Problem 5.R.102c

Textbook Question

Function defined by an integral Let H (𝓍) = ∫₀ˣ √(4 ― t²) dt, for ― 2 ≤ 𝓍 ≤ 2.
(c) Evaluate H '(2) .

Verified step by step guidance

Step 1: Recognize that the function H(𝓍) is defined as an integral, H(𝓍) = ∫₀ˣ √(4 − t²) dt. To find H'(𝓍), we use the Fundamental Theorem of Calculus, which states that if F(𝓍) = ∫ₐˣ f(t) dt, then F'(𝓍) = f(𝓍), provided f is continuous.

Step 2: Apply the Fundamental Theorem of Calculus to H(𝓍). This gives H'(𝓍) = √(4 − 𝓍²), because the integrand √(4 − t²) is continuous for the given domain.

Step 3: Substitute 𝓍 = 2 into the derivative H'(𝓍). This means H'(2) = √(4 − 2²).

Step 4: Simplify the expression inside the square root. Compute 4 − 2², which simplifies to 4 − 4.

Step 5: Conclude that H'(2) = √(0). The derivative at this point is determined by evaluating the square root of the simplified expression.

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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 F is an antiderivative of f on an interval [a, b], then the integral of f from a to b can be computed as F(b) - F(a). This theorem also implies that if H(x) is defined as an integral of a function, then H'(x) can be found by evaluating the integrand at the upper limit of integration.

Differentiation of an Integral Function

When differentiating a function defined by an integral, such as H(x) = ∫₀ˣ f(t) dt, the derivative H'(x) can be computed using the integrand evaluated at the upper limit. Specifically, H'(x) = f(x), provided that f is continuous on the interval. This principle simplifies the process of finding derivatives of integral-defined functions.

Evaluating the Integrand

To evaluate H'(2) in the given problem, we first need to identify the integrand, which is √(4 - t²). We then substitute the upper limit of integration, x = 2, into the integrand. This step is crucial as it allows us to find the value of the derivative at that specific point, which is essential for solving the problem.

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Textbook Question

Max/min of area functions Suppose ƒ is continuous on [0 ,∞) and A(𝓍) is the net area of the region bounded by the graph of ƒ and the t-axis on [0, x]. Show that the local maxima and minima of A occur at the zeros of ƒ. Verify this fact with the function ƒ(𝓍) = 𝓍² - 10𝓍.

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