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

Substitution

Calculus

8. Definite Integrals

Substitution

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2:42 minutes

Problem 8.1.16

Textbook Question

7–64. Integration review Evaluate the following integrals.
16. ∫ from 0 to 1 of (t² / (1 + t⁶)) dt

Verified step by step guidance

Step 1: Recognize that the integral ∫ from 0 to 1 of (t² / (1 + t⁶)) dt involves a rational function. Look for substitution techniques to simplify the integrand.

Step 2: Let u = 1 + t⁶. Then, compute the derivative of u with respect to t: du/dt = 6t⁵. Rearrange to express dt in terms of du: dt = du / (6t⁵).

Step 3: Substitute u and dt into the integral. Replace t² using the substitution u = 1 + t⁶, which implies t² = (u - 1)^(1/3). The integral becomes ∫ (u - 1)^(1/3) / (6t⁵ * u) du.

Step 4: Simplify the expression further by substituting t⁵ in terms of u. From u = 1 + t⁶, t⁵ = ((u - 1)^(5/6)). Replace t⁵ in the denominator and simplify the integrand.

Step 5: Evaluate the resulting integral with respect to u over the transformed limits (from u = 1 to u = 2). Use standard integration techniques or numerical methods if necessary.

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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 calculates the accumulation of a quantity, represented as the area under a curve, between two specified limits. In this case, the integral from 0 to 1 of the function t² / (1 + t⁶) indicates that we are interested in the total area under the curve of this function from t = 0 to t = 1.

Integration Techniques

To evaluate integrals, various techniques can be employed, such as substitution, integration by parts, or recognizing standard forms. For the integral ∫ (t² / (1 + t⁶)) dt, one might consider substitution to simplify the integrand, making it easier to compute the area under the curve.

Continuous Functions

The function being integrated, t² / (1 + t⁶), is continuous over the interval [0, 1]. Continuous functions are essential in calculus because they ensure that the definite integral can be computed without any breaks or discontinuities, allowing for the application of the Fundamental Theorem of Calculus.