Let [tex]g(z)=-\frac{7}{1+z^2}[/tex]
Determine the most general antiderivative of [tex]g[/tex], using [tex]C[/tex] or [tex]c[/tex] as the arbitrary constant, if needed.
[tex]G(z)=\square[/tex]
Answer (2)
The most general antiderivative of the function g(z) = -\frac{7}{1+z^2} is G(z) = -7 \arctan(z) + C, where C is the constant of integration. This is derived from the known antiderivative of \frac{1}{1+z^2}, which is \arctan(z).
;
Recognize that the antiderivative of 1 + z 2 1 is arctan ( z ) .
Multiply the antiderivative by − 7 to get − 7 arctan ( z ) .
Add the constant of integration C .
The most general antiderivative is − 7 arctan ( z ) + C .
Explanation
Problem Analysis We are given the function g ( z ) = − 1 + z 2 7 and we want to find its most general antiderivative G ( z ) . This means we need to find a function whose derivative is g ( z ) .
Recognizing the Antiderivative We know that the derivative of the arctangent function is d z d arctan ( z ) = 1 + z 2 1 . Therefore, the antiderivative of 1 + z 2 1 is arctan ( z ) .
Applying the Constant Factor Since g ( z ) = − 7 ⋅ 1 + z 2 1 , we can find the antiderivative of g ( z ) by multiplying the antiderivative of 1 + z 2 1 by − 7 . This gives us − 7 arctan ( z ) .
Adding the Constant of Integration Finally, we add an arbitrary constant of integration, C , to account for all possible antiderivatives. Thus, the most general antiderivative of g ( z ) is G ( z ) = − 7 arctan ( z ) + C .
Final Answer Therefore, the most general antiderivative of g ( z ) = − 1 + z 2 7 is G ( z ) = − 7 arctan ( z ) + C .
Examples
Antiderivatives are used extensively in physics, such as in determining the position of an object given its velocity function. If the velocity of an object is given by v ( t ) = − 1 + t 2 7 , then the position function s ( t ) is the antiderivative of v ( t ) , which is s ( t ) = − 7 arctan ( t ) + C , where C represents the initial position of the object. This allows us to predict the object's location at any time t .
