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

Previous problemNext problem

3:10 minutes

Problem 8.6.65

Textbook Question

Evaluate the following integrals.
65. ∫ from 0 to 1/6 1/√(1 - 9x²) dx

Verified step by step guidance

Step 1: Recognize the integral's structure. The integrand is of the form

1/√(1 - u²)

, which resembles the derivative of the arcsine function. Specifically, the derivative of

arcsin(u)

1/√(1 - u²)

Step 2: Identify the substitution needed to simplify the integral. Let

u = 3x

, which implies

du = 3 dx

. Rewrite the limits of integration accordingly: when

x = 0

u = 0

, and when

x = 1/6

u = 1/2

Step 3: Substitute into the integral. The integral becomes

∫ from 0 to 1/2 (1/3) * (1/√(1 - u²)) du

, where the factor

1/3

comes from adjusting for

du = 3 dx

Step 4: Apply the antiderivative of

1/√(1 - u²)

, which is

arcsin(u)

. The integral becomes

(1/3) * [arcsin(u)]

, evaluated from

u = 0

u = 1/2

Step 5: Substitute the limits of integration into the result. Compute

(1/3) * [arcsin(1/2) - arcsin(0)]

. Recall that

arcsin(1/2)

and

arcsin(0)

are standard values from trigonometry.

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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 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].

Substitution Method

The substitution method is a technique used to simplify the process of integration by changing the variable of integration. This involves substituting a new variable for a function of the original variable, which can make the integral easier to evaluate. It is particularly useful when dealing with integrals that involve composite functions or complicated expressions.

Trigonometric Substitution

Trigonometric substitution is a specific technique used in integration to simplify integrals involving square roots of quadratic expressions. By substituting a trigonometric function for a variable, such as x = (1/3)sin(θ) for √(1 - 9x²), the integral can often be transformed into a more manageable form. This method leverages the Pythagorean identities to facilitate the integration process.