- #1
escobar147
- 31
- 0
when integrated 1/y^2 becomes -1/y why is this?
daveb said:If you mean integrating with respect to y, try taing the derivative of -1/y.
daveb is asking you to find this derivative:escobar147 said:i still don't understand
The integration of 1/y^2 results in -1/y because of the power rule of integration, which states that the integral of x^n is equal to x^(n+1)/(n+1). In this case, n is equal to -2, so the integral becomes y^(-2+1)/(-2+1) = y^-1/(-1) = -1/y.
The value of -1/y represents the antiderivative or indefinite integral of 1/y^2. In other words, it is the function whose derivative is equal to 1/y^2. This is known as the fundamental theorem of calculus.
The value of y does not have a significant impact on the integration of 1/y^2. The power rule of integration still applies, regardless of the value of y. However, the value of y does affect the overall value of the integral, as seen in the result -1/y.
Yes, the integral of 1/y^2 can be written in a different form using logarithms. By applying the power rule of integration, we get -1/y. This can then be rewritten as ln(y) + C, where C is the constant of integration.
The integration of 1/y^2 has various practical applications in physics, engineering, and economics. For example, it can be used to solve problems involving inverse square laws, such as gravitational force and electric force. It is also useful in determining the total cost of production for a given quantity of goods.