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If \(y^{\prime}=f(x) g(x)\) and if \(g^{\prime}(x)=f(x)\), find \(y . \quad\) Ans. \(\quad y=\frac{[g(x)]^{2}}{2}+C .\)

Short Answer

Expert verified
The function y can be found by integrating y', resulting in \(y = \frac{[g(x)]^2}{2} + C\).

Step by step solution

01

Identify the Relationship Between y', f(x), and g(x)

We are given that the derivative of y with respect to x, denoted as y', is the product of two functions of x: f(x) and g(x). This can be expressed as y' = f(x) g(x). Additionally, we are given that the derivative of g with respect to x, denoted g'(x), is equal to f(x).
02

Recognize the Integral of g'(x)

Since g'(x) is equal to f(x), to find y, we need to integrate y' with respect to x. Based on the relationship y' = f(x)g(x) = g'(x)g(x), we see that we are essentially finding the integral of g'(x) times g(x) with respect to x, which is a product of a function and its derivative.
03

Perform the Integration

The integral of a product of a function and its derivative has the form of \(\int g'(x) g(x) dx = \frac{[g(x)]^2}{2} + C\), where C is the constant of integration. This results from applying the power rule of integration, wherein the integral of \(g'(x)g(x)dx\) is the antiderivative of g(x) raised to the power of 1, which is \(g(x)^2/2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is a fundamental part of mathematics, focused on the process of finding the antiderivatives of functions. This area of calculus is essential because it allows us to calculate the accumulation of quantities, such as area under a curve, total distance traveled, or the volume of an object.

In mathematical terms, the indefinite integral of a function is expressed as \( \int f(x) \, dx \), which represents the collection of all antiderivatives of \( f(x) \) and includes a constant of integration (\( C \)) to cover all possible antiderivatives. Integral calculus deeply connects with differential calculus, as each operation formally undoes the other; differential calculus finds the derivative (rate of change) of a function, while integral calculus finds the original function given the derivative.
Product of Functions
The integration of the product of functions often presents a more complex case in integral calculus. When faced with \( y' = f(x)g(x) \), we are dealing with two intertwined functions. Integrating such products might require special techniques, such as integration by parts or recognizing a special relationship between the functions.

In the context of our example where \( g'(x) = f(x) \), we have such a special case where one function is the derivative of the other. This unique pairing simplifies our efforts because integrating the product of \( g'(x) \) and \( g(x) \) is like reversing the chain rule from differentiation, avoiding the need for more complex integration techniques.
Power Rule of Integration
The power rule of integration is a fundamental technique in integral calculus. It states that for any function \( g(x) \) raised to the power of \( n \) where \( n eq -1 \), the integral is \( \int g(x)^n \, dx = \frac{g(x)^{n+1}}{n+1} + C \). This rule simplifies the integration process by providing a straightforward method to find antiderivatives.

Applying the power rule to our given function \( g(x) \), where \( n = 1 \) and knowing that its derivative \( g'(x) \) equals \( f(x) \), we obtain \( \frac{g(x)^{1+1}}{1+1} + C = \frac{g(x)^2}{2} + C \), clearly illustrating the efficacy of the power rule in determining the integral of \( g'(x)g(x) \).
Constant of Integration
The constant of integration, denoted as \( C \), is a critical part of the indefinite integral that represents an infinite number of possible functions that satisfy the integral equation. Since differentiation of a constant is zero, the antiderivative is not unique, and we include \( C \) to represent all possible vertical translations of the antiderivative.

In practice, \( C \) can be determined if a particular solution is required, for instance, when given an initial condition or boundary value. In our problem where \( y = \int y' \, dx \), we show the constant \( C \) after integrating with the power rule to account for the general solution to the integral of \( g'(x)g(x) \), which encompasses every possible antiderivative.

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