Finding Values for Mean Value Theorem in Integrals

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SUMMARY

The discussion focuses on applying the Mean Value Theorem for Integrals to the function f(x) = x - 2√x over the interval [0, 2]. The integral was calculated as (x^2/2) - (4/3)x^(3/2), leading to the equation f(c)(2-0) = 2 - (8√2)/3. A key insight provided was to substitute u = √c to form a quadratic equation, which aids in finding the values of c, specifically 0.4380 and 1.7908, that satisfy the theorem.

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vee123
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I'm having a some difficulty with this problem:

Fidn the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the indicated interval:

f(x)=x - 2(square root of x) Interval: [0,2]

I found the integral to be: x^2/2 - 4/3 x^3/2, then I solved for the interval and got 2- 8(square root of 2)/3
then I did this:
f(c)(2-0)= 2- 8(square root of 2)/3

I just want to know if I am doing this correctly because I can't seem to get the right answers (0.4380,1.7908)
 
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your methods are correct so far.

you should have an equation that involves c and \sqrt{c}

hint: to solve this equation form the quadratic equation in u, by substituting u = \sqrt{c}

-MS
 
Last edited:
Thanks! I think I know what to do now. I really appreciate it!
 

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