How to Integrate a Fraction with a Square Root in the Denominator

In summary, the conversation is about how to integrate the function (x^5)/((1-2x^3)^(1/2)). One person suggests using the substitution y = \sqrt {1 - 2x^3 } while another person suggests using u = 1 - 2x^3. They discuss the advantages of each substitution and also mention the importance of showing your own thoughts and efforts when seeking help.
  • #1
nyyfan0729
12
0
Can somebody tell me step by step how to integrate (x^5)/((1-2x^3)^(1/2)).
 
Physics news on Phys.org
  • #2
Have you tried the substitution [itex]y = \sqrt {1 - 2x^3 } [/itex]?
 
  • #3
I don't know what you have in mind TD. I would substitute u = 1 - 2x^3, with x^3 = (1 - u)/2
 
  • #4
They're essentially the same thing. But I tend to prefer TD's because it gets rid of all the messy radicals!

nyyfan0729: we're not here to do the problems for you step by step. (that's already done in your textbook and solution manual anyways) We're here to help you when you're stuck: you need to show us your thoughts on the problem, and what you've tried, so that we know where you're stuck and why.
 

FAQ: How to Integrate a Fraction with a Square Root in the Denominator

What is integration in BC Calculus?

Integration in BC Calculus is a mathematical process that involves finding the antiderivative of a given function. It is the inverse operation of differentiation and is used to determine the area under a curve or to find the original function when its derivative is known.

What are the different methods of integration in BC Calculus?

There are several methods of integration in BC Calculus including u-substitution, integration by parts, partial fractions, trigonometric substitution, and integration using tables of integrals. Each method is used depending on the type of function being integrated.

How do you find the area under a curve using integration in BC Calculus?

To find the area under a curve using integration in BC Calculus, you need to first find the antiderivative of the function. Then, you can use the definite integral to calculate the area between the curve and the x-axis within a specific interval. This is represented by the notation ∫f(x)dx, where f(x) is the function and dx represents the infinitesimal change in x.

What is the difference between indefinite and definite integration in BC Calculus?

Indefinite integration in BC Calculus involves finding the antiderivative of a function without specifying limits of integration. It results in a general solution. On the other hand, definite integration involves evaluating the antiderivative within specific boundaries, resulting in a numerical value.

How is integration used in real-life applications?

Integration in BC Calculus has various real-life applications, including calculating the area under a curve in physics and engineering problems, determining the volume of irregular shapes in architecture, and modeling population growth in biology. It is also used in economics to calculate total revenue and in finance to calculate compound interest.

Similar threads

Replies
10
Views
1K
Replies
15
Views
1K
Replies
6
Views
575
Replies
47
Views
2K
Replies
16
Views
2K
Replies
2
Views
1K
Replies
5
Views
941
Back
Top