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

12. Techniques of Integration

Improper Integrals

Calculus

12. Techniques of Integration

Improper Integrals

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Multiple Choice

Which of the following integrals is improper because the interval of integration is infinite?

\int_{0}^{1} x^{2} d x

\int_{1}^{2} \sqrt{x} d x

\int_{1}^{\infty} \frac{1}{x^{2}} d x

\int_{- 1}^{1} \frac{1}{x} d x

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Verified step by step guidance

Step 1: Understand the definition of an improper integral. An integral is considered improper if it involves an infinite interval of integration or if the integrand becomes unbounded within the interval.

Step 2: Analyze each integral provided in the problem. For example, int_0^1 x^2 dx has a finite interval [0, 1] and the integrand x^2 is well-behaved (bounded) within this interval, so it is not improper.

Step 3: Examine int_1^2 sqrt{x} dx. The interval [1, 2] is finite, and the integrand sqrt{x} is also bounded within this interval, so this integral is not improper.

Step 4: Consider int_1^{infty} frac{1}{x^2} dx. The interval of integration is [1, ∞), which is infinite. This makes the integral improper due to the infinite upper limit of integration.

Step 5: Review int_{-1}^1 frac{1}{x} dx. The interval [-1, 1] is finite, but the integrand frac{1}{x} becomes unbounded at x = 0 (division by zero). This makes the integral improper due to the unbounded behavior of the integrand within the interval.