Let C be the curve parameterized by , for . Find the value of the line integral .
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
Introduction to Definite Integrals
Problem 8.1.12
Textbook Question
7–64. Integration review Evaluate the following integrals.
12. ∫ from -5 to 0 of dx / √(4 - x)

1
Step 1: Recognize the integral ∫ dx / √(4 - x) as a definite integral with limits of integration from -5 to 0. The integrand involves a square root, which suggests a substitution method might be helpful.
Step 2: Let u = 4 - x. Then, differentiate u with respect to x to find du/dx = -1, or equivalently, dx = -du. Substitute u into the integral to replace x and dx.
Step 3: Adjust the limits of integration according to the substitution. When x = -5, u = 4 - (-5) = 9. When x = 0, u = 4 - 0 = 4. The integral now becomes ∫ from u=9 to u=4 of -du / √u.
Step 4: Simplify the integral using the negative sign and rewrite it as -∫ from u=9 to u=4 of du / √u. Recognize that 1/√u can be expressed as u^(-1/2), which is a standard power rule for integration.
Step 5: Apply the power rule for integration to u^(-1/2). The integral of u^(-1/2) is 2√u. Evaluate this antiderivative at the new limits of integration (u=9 and u=4), and subtract the results to find the value of the definite integral.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral represents the signed area under a curve between two specified limits. It is denoted as ∫ from a to b of f(x) dx, where 'a' and 'b' are the lower and upper limits, respectively. The result of a definite integral is a numerical value that quantifies the accumulation of the function's values over the interval [a, b].
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Integration Techniques
Integration techniques are methods used to find the integral of a function. Common techniques include substitution, integration by parts, and recognizing standard forms. For the integral ∫ dx / √(4 - x), a substitution method can simplify the expression, making it easier to evaluate the integral.
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Square Root Functions
Square root functions, such as √(4 - x), are important in calculus as they often appear in integrals and derivatives. Understanding their behavior, including their domain and range, is crucial for evaluating integrals involving them. In this case, the expression under the square root must be non-negative, which influences the limits of integration.
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