cristo said:Which substitutions have you tried? What about u=x-2?
Note in future that you should post these sort of questions in the suitable homework forum, and no the technical forums.
ice109 said:how would that work?
cristo said:Let u=x-2, then du=dx and x=u+2. This converts the integral into [tex]\int\frac{5(u+2)du}{u}=\int\left[5+\frac{10}{u}\right]du[/tex]
Integration is a mathematical process that involves finding the area under a curve on a graph. It is the inverse of differentiation and is used to calculate the total change in a quantity over a given interval.
The formula for integration is ∫f(x)dx, where f(x) represents the function being integrated and dx represents the infinitesimal change in x.
To integrate (5x+2)dx/(x-2) from 0->1, you can use the substitution method. Let u = x-2, then du = dx. The integral becomes ∫(5u+12)du/u from -2->-1. Using the power rule for integration, the integral becomes 5ln(u) + 12ln(u) from -2->-1. Substituting back for u, the final answer is 5ln(x-2) + 12ln(x-2) from 0->1.
The limits of integration (0->1) represent the interval over which the integration is being performed. In this case, it means that the area under the curve of the function (5x+2)dx/(x-2) is being calculated from x=0 to x=1.
Yes, integration is used in various fields such as physics, engineering, economics, and more to solve problems related to rates of change, areas, volumes, and other quantities. It is a powerful tool in understanding and analyzing real-world phenomena.