ruiwp13 said:Homework Statement
∫x^3*√(x^2-5) dx
Homework Equations
∫u.dv=u.v-∫du.v
The Attempt at a Solution
So i tried to change the integral to ∫x*x^2*√(x^2-5)dx and u = x^2-5, then du = 2x, so 1/2*∫x^2*√(x^2-5) . Let u = √(x^2-5) , du = x/√(x^2-5) and dv = x^2 , v = x^3/3. Am I going in the right direction?
Thanks in advance
Dick said:You don't need parts. Your first substitution is the one to use, u = x^2-5. Notice that means x^2=5+u.
ruiwp13 said:so I multiplied and I got : 1/2*∫u^(3/2)+5u^(1/2)
1/2*1/5 * ∫u^(3/2)+u^(1/2)
Dick said:Just sloppy algebra, I think. Where did the '1/5' come from?
ruiwp13 said:To remove the 5 from 5u^(1/2)
The solution is supposed to be:
x^3/3*(u^(3/2))-2/15*(u^(5/2))+c
Dick said:'Remove the 5'?? I don't get it. And the given solution looks wrong as well.
ruiwp13 said:
Dick said:Wolfram is using u^(5/2)=u*u^(3/2) and pulling out the common factor of u^(3/2).
ruiwp13 said:So where is my mistake?
The purpose of finding the integral of ∫x^3*√(x^2-5) dx is to determine the area under the curve of the function and to evaluate the function for different values of x.
The integral of ∫x^3*√(x^2-5) dx can be solved using the substitution method or integration by parts. It involves breaking down the function into simpler forms and applying integration rules to evaluate the integral.
Common mistakes to avoid when solving for the integral of ∫x^3*√(x^2-5) dx include incorrect use of integration rules, forgetting to add the constant of integration, and making errors in the substitution or integration by parts process.
Yes, the integral of ∫x^3*√(x^2-5) dx is only defined for values of x that make the function continuous and differentiable. This means that the domain of the function is restricted to x values that make the expression inside the square root greater than or equal to 0.
The integral of ∫x^3*√(x^2-5) dx has various real-life applications, such as calculating the work done by a variable force, determining the center of mass for a given shape, and evaluating the total distance traveled by an object with varying velocity. It is also used in engineering, physics, and economics to solve various problems related to rates of change and optimization.