Questions: Find the indefinite integral. Check your result by differentiating. (Use C for the constant) ∫ v^(-1 / 2) dv

Solution Steps

To find the indefinite integral of \( v^{-1/2} \), we can use the power rule for integration. The power rule states that the integral of \( v^n \) is \( \frac{v^{n+1}}{n+1} \) plus a constant of integration \( C \), provided \( n \neq -1 \). After integrating, we can differentiate the result to check our work.

1. Apply the power rule for integration to \( v^{-1/2} \).
2. Simplify the resulting expression.
3. Add the constant of integration \( C \).
4. Differentiate the result to verify the correctness.

To find the indefinite integral of \( \int v^{-1/2} \, dv \), we apply the power rule for integration. The integral of \( v^n \) is given by:

\[ \int v^n \, dv = \frac{v^{n+1}}{n+1} + C \]

In this case, \( n = -\frac{1}{2} \), so:

\[ \int v^{-1/2} \, dv = \frac{v^{(-1/2) + 1}}{(-1/2) + 1} + C = \frac{v^{1/2}}{1/2} + C = 2v^{1/2} + C \]

The indefinite integral simplifies to:

\[ \int v^{-1/2} \, dv = 2\sqrt{v} + C \]

To verify our result, we differentiate \( 2\sqrt{v} + C \):

\[ \frac{d}{dv}(2\sqrt{v} + C) = 2 \cdot \frac{1}{2} v^{-1/2} = \frac{1}{\sqrt{v}} \]

This confirms that the derivative of our integral matches the original integrand \( v^{-1/2} \).

Final Answer