Solving the Integral of s*(4-s)^\frac{1}{2}

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In summary, the conversation discusses finding the integral of s*(4-s)^\frac{1}{2} using u-substitution and making a substitution of u=4-s. The resulting integral is 4u^\frac{1}{2}*u^\frac{3}{2} and the final solution involves plugging in 4-x for u. There is also a clarification about the original integral being meant for ds and not dx, and a correction for the value of du.
  • #1
yaho8888
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Homework Statement



Find the intergal of [tex] s*(4-s)^\frac{1}{2} [/tex]

Homework Equations



[tex] \int s*(4-s)^\frac{1}{2} dx [/tex]

The Attempt at a Solution



using u sub
u = 4-s
s = 4-u

[tex] \int (4-u)*(u)^\frac{1}{2} du [/tex]

[tex] \int 4u^\frac{1}{2}*u^\frac{3}{2} [/tex]

then what?
 
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  • #2
You mean, [tex] \int (4u^\frac{1}{2}-u^\frac{3}{2})du [/tex]
 
  • #3
Yes, [tex] \int (4u^\frac{1}{2}-u^\frac{3}{2})du [/tex]

to solve this do I just do the integral of the equation above then plug in 4-x for u?
 
  • #4
yes, 4-s. I think your original integral was meant to be ds not dx right?

also when you made your u substitution du=-ds so your integral should be u^(3/2)-4*u^(1/2)
 

FAQ: Solving the Integral of s*(4-s)^\frac{1}{2}

What is the formula for solving the integral of s*(4-s)^\frac{1}{2}?

The formula for solving this integral is ∫ s*(4-s)^\frac{1}{2} ds = 2/3 * (4-s)^\frac{3}{2} + C, where C is the constant of integration.

How do you determine the limits of integration for this integral?

The limits of integration can be determined by setting the expression inside the integral equal to 0 and solving for s. In this case, the limits are 0 and 4.

What is the general strategy for solving integrals with square roots?

The general strategy for solving integrals with square roots is to use the substitution method. This involves substituting a new variable for the expression inside the square root, and then solving the integral using the new variable.

Can this integral be solved without using the substitution method?

Yes, this integral can also be solved using the power rule, where the exponent of the variable is increased by 1 and then divided by the new exponent. However, the substitution method is typically more efficient for integrals with square roots.

Are there any special cases for solving this integral?

Yes, if the expression inside the square root is a perfect square, then the integral can be solved using the trigonometric substitution method. This involves substituting trigonometric functions for the variable and then using trigonometric identities to simplify the integral.

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