INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 0 answersPost by chandler corbeil on July 13, 2014When 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.