<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>If f(x) is differentiable on the open interval (a,b) and further continuous on the closed interval [a,b] ==> Then there exists at least one point c in (a,b) such that</p>
<p>f'(c) = f(b) -f(a) / (b-a)</p>
<p>So, f(x) = x -2x^1/2 and f'(x) = 1 - 1/x^1/2</p>
<p>substitute f(2) = 2 - 2 sqrt(2), f(0) =0</p>
<p>and f'(c) = 1 - 1/c^1/2 in to the theorem</p>
<p>This simply yields c = 1/2</p>
<p>i think you're doing the mean value theorem for derivatives.</p>
<p>Math teacher made great review page that includes this and more on a pdf here: <a href="http://siths.org/ourpages/auto/2007/2/27/1172603480261/Formulas%20and%20Theorems-2007.doc?rn=5076014%5B/url%5D">http://siths.org/ourpages/auto/2007/2/27/1172603480261/Formulas%20and%20Theorems-2007.doc?rn=5076014</a></p>
<p>if f is continuous on [a,b], then there exists a number c in [a,b] such that</p>
<p>f(x) dx on the integral from a to b = f(c)(b-a) </p>
<p>hope this helps. just took my calc final.</p>
<p>i know the equation but how would you solve it??? my text book has Two decimal answers at the back of the book, but i have no clue how they got those answers.</p>
<p>f(x)=x-2x^1/2, interval= (0,2)
so b-a is 2-0 = 2
F(x) = x^2/2 - 2x^(3/2) / (3/2) = x^2/2 - 4/3<em>x^3/2 now do at 2 and 0
at 2 F(2) is [2-4/3</em>sqrt(8)] and F(0) is 8.
So now you set f(c) = [F(2)-F(0)] / [2-0] and solve for c</p>
<p>What this actually means.
The integral will give you an area. b-a will give you a length. This is saying that if the function is continuous then for some point along the interval, from a to b, the function must take on the average height. AKA make the integral into a rectangle with base (b-a) and some height which will be the average height. At some point the function if it is continuous must take on that height, you are looking for the x value at which it does.</p>
<p>thats the same answer that i got but it seems like the books answer is different. I'll ask my teacher tomorrow. also DOESN't f(0)=0 not 8</p>
<p>but can we expand the above theorem into the integral form ?</p>
<p>if we define, F(x) = int (f(t), t= x,a) </p>
<p>Then F'(x) = f(x). And now, substitute that into the above theorem, then we get</p>
<p>F'(c) = F(b) - F(a) / (b-a)</p>
<p>which is simply as </p>
<p>int (f(t), t= b,a) = f (c)*(b-a)</p>
<p>SO, again substitute b = 2, a =0
and solve for c !! my maple says</p>
<p>c = 1/9<em>(3+2sqrt(3) -sqrt(6))</em><em>2 or 1/9</em>(-3+2sqrt(3)-sqrt(6))**2</p>
<p>make sure the answer falls in the interval (0,2) if not==> dismiss one with extreme prejudice</p>
<p>ok here are details:</p>
<p>F(x) = 1/2 x^2 -4/3x^3/2 + const</p>
<p>LHS :F(2) -F(0) = 2-8/3<em>sqrt(2)
RHS: 2(c-2</em>sqrt(c))</p>
<p>let sqrt(c) = C and solve for C in quadratic eqs.</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>