<p>Find the value of c guaranteed by the mean value theorum for Integrals for the function over the Specified interval ???</p>
<p>f(x)=x-2x^1/2, interval= (0,2)</p>
<p>I am great at calculus but i just dont understand this theorem. Anyone wants to explain please??? thanks</p>
<p>Anybody???</p>
<p>i don't know if this is right, but is the MVT for integrals just the average value of the function over that integral? maybe?</p>
<p>so (1/2) fint(x-2x^1/2, x, 0, 2) </p>
<p>haha not really that familiar with ti83 keystroks anymore...</p>
<p>but anyone wanna verify? not 100% sure about the c thing, but i do know that mvt-integrals helps us do the average value. we only did the c value for mvt derivatives.</p>
<p>There are two answers at the back of my book but they are totally different from my answer and Math 9 Ti-84 answer HOW DID THEY GET TWO ANSWERS?</p>
<p>The mean value theorem:</p>
<p>(f(b)-f(a))/(b-a) = d(f(x))/dx</p>
<p>so i get .5,0,2 as possible values</p>
<p>well, see that's what i thought at first, but i'm pretty sure that's mean value theorem for derivatives. and i know that mvt-integrals guarantees the existence the average value.</p>
<p>hmm... i know how to find average value
1/b-a antiderivative of f(x)dx at a and b. (a,b)</p>
<p>If u integrate the function
f(x)=x-2x^1/2, interval= (0,2)
the answer is approximately -1.77124.</p>
<p>The MVT for integrals says that there is at least one value of c such that [the integral of function] = f(c)(b-a).</p>
<p>So we just look for the value of c. (b-a) is 2, so f(c)(2) = -1.77124.
Divide both sides by 2, and f(c) = -.88562, or c-2c^(1/2) = -.88562</p>
<p>Solve for c and u get c = .4380 or 1.7908 (approximate values).</p>
<p>I hope this is correct...</p>
<p>i asked my teacher and yes thats exactly the way she told me.</p>
<p>there's another one, i finished part of it but how do i make it into quadratic eqaution:</p>
<p>f(x)=9/x^3 [1,3}</p>
<p>i know that would equal 2. </p>
<p>but after that 2=9/c^3
2-9/c^3=0 can i apply the quadratic formula over here</p>
<p>NEVER MIND THAT, I FOUND THE ANSWER!!</p>
<p>but what about f(x)=2sec^2x [-pi/4,pi/4]</p>
<p>i have solved the equation part of it which is 8/pi or 2.5465</p>
<p>SO WHATS: 8/pi=2sec^2c(since im solving for c)</p>
<p>SO WHATS: 8/pi=2sec^2c(since im solving for c)</p>
<p>I think it's arccos(sqrt(pi)/2) if I solved it correctly</p>