Lo.Lee.Ta. said:∫x^2(√(x^3 + 1))
u = x^3 + 1
du = 3x^2dx
(1/3)du = x^2dx
∫√(u)*(1/3)du
= 2/3(u)^3/2 *(1/3)
= 2/9(x^3 + 1)^3/2
Is this the correct answer? Thank you.
Lo.Lee.Ta. said:Thanks, you guys! :)
That's true. I should differentiate when I'm not sure...
The integral of (x^2)(sqrt(x^3 + 1))dx represents the area under the curve of the function (x^2)(sqrt(x^3 + 1)) between the limits of integration.
To solve the integral of (x^2)(sqrt(x^3 + 1))dx, you can use the substitution method by setting u = x^3 + 1 and du = 3x^2 dx. This will transform the integral into ∫ (1/3)(sqrt(u)) du, which can be easily solved using the power rule and basic integration techniques.
Yes, the integral of (x^2)(sqrt(x^3 + 1))dx can be solved using the power rule after applying the substitution method to simplify the integral.
The limits of integration for the integral of (x^2)(sqrt(x^3 + 1))dx depend on the specific problem or scenario given. However, they typically involve the values of x that define the region under the curve of the function.
The integral of (x^2)(sqrt(x^3 + 1))dx has various real-world applications in fields such as physics, engineering, and economics. It can be used to calculate the volume of irregularly shaped objects, the work done by a variable force, and the area under a demand curve in economics, among others.