If F1 15 F is Continuous and 3 F X Dx 1 15 What is the Value of F3
Mean Value Theorem homework
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If [tex] f [/tex] is continuous and [tex] \int^{3}_{1} f(x) dx = 8 [/tex], show that [tex] f [/tex] takes on the value 4 at least once on the interval [1,3] . I know that the average value of [tex] f(x) [/tex] is 4. So does this imply that [tex] f_{ave} = f(c) = 4 [/tex] and [tex] f(x) [/tex] takes on the value of 4 at least once on the interval [1,3] ?
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I'd go with contradiction. Suppose there is no x0 in [1,3] such that f(x0)=4. With that assumption, what does the mean value thm says about the values f can take on [1,3]? There are two cases, etc.
Yes, but not directly--the mean value theorem is about derivatives. What is the derivative of g(x) = [tex] \int_{3}^{x} f(t) dt [/tex]?If [tex] f [/tex] is continuous and [tex] \int^{3}_{1} f(x) dx = 8 [/tex], show that [tex] f [/tex] takes on the value 4 at least once on the interval [1,3] . I know that the average value of [tex] f(x) [/tex] is 4. So does this imply that [tex] f_{ave} = f(c) = 4 [/tex] and [tex] f(x) [/tex] takes on the value of 4 at least once on the interval [1,3] ?
I'd go with contradiction. Suppose there is no x0 in [1,3] such that f(x0)=4. With that assumption, what does the mean value thm says about the values f can take on [1,3]? There are two cases, etc.
make that the intermediate value thm, sorry.
Anyway, courtigrad's way is a little bit faster, more elegant and "on topic" provided this is an exercice designed to make you apply the mean valut thm.
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Thanks for your help. Lets say I want to prove (not rigorously) the Mean Value Theorem for Integrals, [tex] \int_{a}^{b} f(x)\dx = f(c)(b-a) [/tex] by using the Mean Value Theorem for Derivatives to the function [tex] F(x) = \int^{x}_{a} f(t)\dt [/tex]. We know that [tex] F(x) [/tex] is continous on [a,x] and differentiable on (a,x) . So [tex] F(x) - F(a) = f(c)(x-a) [/tex] or [tex] \frac{\int^{b}_{a} f(t)\dt - \int_{a}^{a} f(t)\dt}{b-a} = f(c)(b-a) [/tex]. So [tex] f(c) = \frac{1}{b-a}\int_{a}^{b} f(t)\dt [/tex]. Is this correct?
Thanks
minus the extra (b-a) in the second to last equation, I'd say this is a perfectly acceptable and rigourous proof of the mean value thm for integral. Nice job!
This step
[tex] \frac{\int^{b}_{a} f(t)\dt - \int_{a}^{a} f(t)\dt}{b-a} = f(c)(b-a) [/tex]
is not true. Also instead of just writing equations, you should explain what you're doing--for example you should say, "where c is some value between a and b" and you should say when you use the mean value theorem or fundamental theorem of calculus to justify a step.
[tex] \frac{\int^{b}_{a} f(t)\dt - \int_{a}^{a} f(t)\dt}{b-a} = f(c)(b-a) [/tex]
is not true. Also instead of just writing equations, you should explain what you're doing--for example you should say, "where c is some value between a and b" and you should say when you use the mean value theorem or fundamental theorem of calculus to justify a step.
Because of the way you've presented it, I'm not sure if you are already aware of this, but what is d(F(x))/dx? What is F(a) and F(b)?
[tex] \frac{dF(x)}{dx} = f(t) [/tex]. [tex] F(a) = 0 [/tex] and [tex] F(b) = \int_{a}^{b} f(t) [/tex]
(you rock!) 
should read "We know by one version of the fond thm of calculus that under the assumption of the continuity of f on [a,b], [tex] F(x) [/tex] is continous on [a,b] , differentiable on (a,b) and that F'(x)=f(x). Therefor, by the mean value thm, there exists a number c in (a,b) such that F(b)-F(a)=F'(c)(b-a). But F'(c) = f(c), [itex]F(b)=\int_a^b f(t)dt[/itex], F(a)=[itex]\int_a^a f(t)dt=0[/itex], hence the result."
But I agree with ortho on the last part: in an exam, you'd most certainly lose points for not expliciting your reasoning in terms of the condition of applicability of the thm, etc. For instance, the step that goes,
We know that [tex] F(x) [/tex] is continous on [a,x] and differentiable on (a,x) . So [tex] F(x) - F(a) = f(c)(x-a) [/tex]
should read "We know by one version of the fond thm of calculus that under the assumption of the continuity of f on [a,b], [tex] F(x) [/tex] is continous on [a,b] , differentiable on (a,b) and that F'(x)=f(x). Therefor, by the mean value thm, there exists a number c in (a,b) such that F(b)-F(a)=F'(c)(b-a). But F'(c) = f(c), [itex]F(b)=\int_a^b f(t)dt[/itex], F(a)=[itex]\int_a^a f(t)dt=0[/itex], hence the result."
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Let [itex]F(x)= \int_1^x f(t)dt[/itex]. Then [itex]F(1)= \int_1^1 f(t)dt= 0[/itex] and [itex]F(3)= \int_1^3 f(t)dt= 8[/itex] so [itex]\frac{F(3)- F(1)}{3- 1}= \frac{8- 0}{2}= 4[/itex]. By the mean value theorem then, there must exist c between 1 and 3 such that F'(c)= f(c)= 4.
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