Discover How to Solve the Integral of 1/[9.8 - (1/245)v^2] Quickly

Thread starter ns5032
Start date Feb 17, 2008
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Integral

In summary, the integral of 1/[9.8 - (1/245)v^2] can be quickly solved using the substitution method, by letting u = 9.8 - (1/245)v^2. The number 9.8 in the integral represents the acceleration due to gravity on Earth. While it is possible to solve the integral using partial fraction decomposition, substitution is a quicker method. This integral can be applied to any real value of v, but is commonly used in physics problems involving velocity. It has practical applications in calculating work and potential energy.

Feb 17, 2008

ns5032

How do I find the integral of:

1/[9.8 - (1/245)v^2]

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Feb 17, 2008

Dick

Science Advisor

Homework Helper

Write the denominator as the difference of two squares and factor it. As in a^2-b^2=(a-b)*(a+b). Then use partial fractions.

1. How can I solve the integral of 1/[9.8 - (1/245)v^2] quickly?

The most efficient way to solve this integral is by using the substitution method. Let u = 9.8 - (1/245)v^2 and du = (-2/245)v dv. This will simplify the integral to ∫1/u du, which is a simple logarithmic function.

2. What is the significance of the number 9.8 in the integral?

The number 9.8 represents the acceleration due to gravity on Earth in m/s^2. This integral is commonly used in physics problems involving gravitational force.

3. Can this integral be solved without using the substitution method?

Yes, it is possible to solve this integral using partial fraction decomposition. However, this method may not be as quick as using substitution.

4. Is there a specific range of values for v that this integral can be applied to?

This integral can be applied to any value of v, as long as it is a real number. However, it is commonly used for values of v that represent velocities in physics problems.

5. Are there any practical applications for this integral?

Yes, this integral has practical applications in physics, specifically in calculating the work done by a varying gravitational force. It can also be used to calculate the potential energy of an object in motion.