integration of (qdq) please find my answer with a fully explained

To tackle the integration of the function \( qdq \), we first need to clarify what \( q \) and \( dq \) represent. In calculus, \( dq \) typically denotes a differential element related to the variable \( q \). Therefore, the expression can be interpreted as integrating a function of \( q \) with respect to \( q \). Let’s break this down step by step.

Understanding the Function

Assuming \( q \) is a variable, the expression \( qdq \) can be rewritten for integration purposes. The term \( qdq \) suggests that we are looking at the function \( q \) multiplied by its differential \( dq \). This can be viewed as a simple integration problem.

Setting Up the Integral

The integral we want to solve is:

\( \int q \, dq \)

Applying the Power Rule

To integrate \( q \, dq \), we can use the power rule of integration. The power rule states that:

• If \( f(q) = q^n \), then \( \int f(q) \, dq = \frac{q^{n+1}}{n+1} + C \) for \( n \neq -1 \).

In our case, we can consider \( q \) as \( q^1 \). Thus, applying the power rule:

\( \int q \, dq = \frac{q^{1+1}}{1+1} + C = \frac{q^2}{2} + C \)

Final Result

So, the integral of \( qdq \) is:

\( \frac{q^2}{2} + C \)

Here, \( C \) represents the constant of integration, which accounts for any constant value that could have been present before differentiation.

Example for Clarity

Let’s consider a practical example. If we want to find the area under the curve of the function \( q \) from \( q = 0 \) to \( q = 3 \), we would evaluate:

\( \int_0^3 q \, dq = \left[ \frac{q^2}{2} \right]_0^3 = \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2} = 4.5 \)

This result indicates that the area under the curve from \( q = 0 \) to \( q = 3 \) is 4.5 square units.

Conclusion

In summary, integrating \( qdq \) leads us to the expression \( \frac{q^2}{2} + C \). This fundamental technique is essential in calculus and can be applied to various functions, enhancing our understanding of areas, volumes, and other applications in mathematics and physics.