Sign In | Subscribe
INSTRUCTORS Raffi Hovasapian John Zhu
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Calculus AB
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Related Books

Bookmark and Share
Lecture Comments (1)

0 answers

Post by chandler corbeil on July 13, 2014

When is he going to be teaching multivariable calculus. he's the easiest to pay attention to.

Integrals of Power Rule

  • To avoid confusion, treat n as just a constant
  • Misleading constants: natural logs, fractions, and decimals
  • Careful not to confuse the power rule of integrals with the power rule of derivatives

Integrals of Power Rule

∫x  dx
  • ∫x dx = ∫x1 dx
  • ∫x1 dx = [(x1 + 1)/(1 + 1)] + c
  • where c is some constant. By convention, c or C typically denotes a constant when working with integrals.
[(x2)/2] + c
∫x2 dx
  • ∫x2 dx = [(x2 + 1)/(2 + 1)] + c
∫x2 dx = [(x3)/3] + c
∫x10 dx
  • ∫x10 dx = [(x10 + 1)/(10 + 1)] + c
∫x10 = [1/11]x11 + c
Find [d/dx] ([1/11] x11 + c) where c is some constant
  • [d/dx] ([1/11] x11 + c) = [1/11] [d/dx] x11 + [d/dx] c
  • [d/dx] ([1/11] x11 + c) = [11/11] x10 + 0
[d/dx] ([1/11] x11 + c) = x10
∫1  dx
  • ∫1  dx = ∫x0 dx
  • ∫1  dx = [(x0 + 1)/(0 + 1)] + c
  • ∫1  dx = [(x1)/1] + c
∫1  dx = x + c
∫[1/(x2)] dx
  • ∫[1/(x2)] dx = ∫x−2 dx
  • ∫[1/(x2)] dx = [(x−2 + 1)/(−2 + 1)] + c
  • ∫[1/(x2)] dx = [(x−1)/(−1)] + c
∫[1/(x2)] dx = −[1/x] + c
∫x−[3/2] dx
  • ∫x−[3/2] dx = [(x−[3/2] + 1)/(−[3/2] + 1)] + c
  • ∫x−[3/2] dx = [1/(−[1/2])] x−[1/2] + c
∫x−[3/2] dx = −2 [1/(√x)] + c
5∫x−[3/2] dx
  • We know from the previous problem that ∫x−[3/2] dx = −2 [1/(√x)] + c
  • When there is a constant in front of an integral, all it is is simple multiplication
  • 5∫x−[3/2] dx = 5 (−2 [1/(√x)] + c)
  • 5∫x−[3/2] dx = −10[1/(√x)] + c1
  • What we actually name the constant is irrelevant. I called it c1 to differentiate from the other c we used earlier, which aren't equal because of the 5
5∫x−[3/2] dx = −10[1/(√x)] + c1
∫[(x[3/2])/(x3)] dx
  • ∫[(x[3/2])/(x3)] dx = ∫x[3/2] x−3 dx
  • ∫[(x[3/2])/(x3)] dx = ∫x[3/2] − 3 dx
  • ∫[(x[3/2])/(x3)] dx = ∫x−[1/2]
  • ∫[(x[3/2])/(x3)] dx = [(x−[1/2] + 1)/(−[1/2] + 1)] + c
  • ∫[(x[3/2])/(x3)] dx = [1/([1/2])] x[1/2] + c
∫[(x[3/2])/(x3)] dx = 2 √x + c
∫[1/x] dx
  • ∫[1/x] dx = ∫x−1 dx
  • ∫[1/x] dx = [(x−1 + 1)/(−1 + 0)] + c = [1/0] + c
  • Undefined! The integral product rule cannot be used on [1/x]
  • We can work backwards
  • The derivative of what function results in [1/x]?
  • [d/dx] (lnx + c) = [1/x]
∫[1/x] dx = lnx + c

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Integrals of Power Rule

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Power Rule 0:06
    • Example 1
  • Example 2 2:02
  • Example 3 2:54
  • Example 4 3:45
  • Example 5 4:49
  • Example 6 6:47