The phrase "integration of dV V" can be interpreted in a couple of ways due to its unconventional notation. To provide a comprehensive and exact answer, we will address the two most probable interpretations in integral calculus.
Understanding the Integration of dV V
The way an integral is written typically involves the function to be integrated followed by the differential of the variable of integration (e.g., $\int f(V) dV$). The notation "dV V" can be seen as either $V$ multiplied by $dV$, or potentially a typographical error for $1/V$ multiplied by $dV$.
Interpretation 1: Integration of V with respect to V ($\int V \, dV$)
This is the most direct and common interpretation when 'dV' signifies the differential and 'V' is the function being integrated. In this case, it follows the power rule of integration.
The Power Rule of Integration
The power rule states that for any real number $n \ne -1$:
$$ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $$
where $C$ is the constant of integration.
Applying this rule to $\int V \, dV$:
- Here, $V$ can be written as $V^1$. So, $n=1$.
- Adding 1 to the exponent gives $1+1=2$.
- Dividing by the new exponent gives $1/2$.
Therefore, the exact answer for this interpretation is:
$$ \int V \, dV = \frac{V^{1+1}}{1+1} + C = \frac{V^2}{2} + C $$
Example:
If you need to find the integral of $V$ from $V=1$ to $V=3$:
$$ \int_1^3 V \, dV = \left[ \frac{V^2}{2} \right]_1^3 = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4 $$
Interpretation 2: Integration of 1/V with respect to V ($\int \frac{1}{V} \, dV$)
Sometimes, the phrasing "dV V" might be a concise way of saying "dV over V" or a typo for "dV/V". This interpretation is especially relevant given that it addresses a fundamental and distinct integration rule.
The Logarithmic Rule of Integration
When the power rule cannot be applied (specifically when $n=-1$), the integral of $x^{-1}$ (or $1/x$) is the natural logarithm of the absolute value of $x$.
A well-known result in calculus states that the integral of $\frac{1}{V}$ with respect to $V$ is the natural logarithm of the absolute value of $V$, plus a constant of integration. For instance, the indefinite integral of dv/v is expressed as ln(v) + c.
Therefore, the exact answer for this interpretation is:
$$ \int \frac{1}{V} \, dV = \ln|V| + C $$
The absolute value is crucial because the natural logarithm is only defined for positive numbers, but $V$ can be negative.
Definite Integrals for 1/V
When evaluating this type of integral over a specific range, such as from $V_1$ to $V_2$, it becomes a definite integral. This is calculated by taking the difference of the natural logarithm evaluated at the upper and lower limits:
$$ \int_{V_1}^{V2} \frac{1}{V} \, dV = [\ln|V|]{V_1}^{V_2} = \ln|V_2| - \ln|V_1| $$
This approach is crucial in many physics and engineering applications, such as calculating work done by a gas during isothermal expansion or analyzing RC circuits.
Example:
If you need to find the integral of $1/V$ from $V=1$ to $V=e$:
$$ \int_1^e \frac{1}{V} \, dV = [\ln|V|]_1^e = \ln|e| - \ln|1| = 1 - 0 = 1 $$
Summary of Integrals
To summarize the two interpretations:
| Interpretation | Integral Form | Result | Conditions |
|---|---|---|---|
| Standard Power Rule | $\int V \, dV$ | $\frac{V^2}{2} + C$ | Applies generally |
| Reciprocal Function | $\int \frac{1}{V} \, dV$ | $\ln | V |
Understanding which integral is intended relies on the precise mathematical context or notation used. Both are fundamental results in integral calculus.
