91Ó°ÊÓ

In calculus, integrating a derivative helps us find the original function. Given the derivative, we can work backwards through integration. In the exercise provided, the derivative of our function is:

\(f^{\prime}(x) = x\)
When we integrate this, we are essentially accumulating the area under the curve of the derivative, which brings us back to the original function. For this purpose, we set up the integral as follows:

\[f(x) = \int x \: dx = \frac{1}{2} x^2 + C \]

Notice the addition of the constant \(C\). This constant appears because when we differentiate back from \(f(x)\), any constant would disappear, as shown in the differentiation table. This constant represents all possible vertical shifts of the curve of \(f(x)\).

Initial conditions are given specific points that a function must pass through. These are crucial in determining the constant of integration. In our problem, the initial condition is:

\(f(0) = 3\)

This means that when \(x = 0\), the value of the function must be 3. We use this condition to solve for \(C\), the constant of integration in our integrated function. From our previous step, we have:

\[f(x) = \frac{1}{2} x^2 + C\]

Plugging in the initial condition, we get:

\[f(0) = \frac{1}{2} (0)^2 + C = 3\]

This simplifies to:

\[C = 3\]

Using this \(C\) value, we can now write the definite function:

\[f(x) = \frac{1}{2} x^2 + 3\]

Integration is a fundamental concept in calculus, and knowing how to integrate is crucial for solving many calculus problems. There are different techniques of integration, including:

• Basic Integration: Integrating simple functions like \(x\), which we used in the problem.
• Integration by Substitution: Useful for functions that are products of a function and its derivative.
• Integration by Parts: Used when integrating the product of two functions.

In the given exercise, only basic integration is needed. We integrate \(x\) by recognizing that the antiderivative of \(x\) is \(\frac{1}{2} x^2\). Practices like these form the foundation for more complex problems involving more elaborate techniques.