1. ## Final Calculus Probset

Can you help me solve my problem set please? I already solved no 3 and am working on the others. but as of now, finishing it is far and deadline's on monday. Really need to finish this all but can't do it without help from you guys. please help me. thanks!

or see attachments below

2. Can you at least show some work and your own thoughts on these problems? It'd be easier for us to point you in the direction than starting right from scratch.

3. Originally Posted by o_O
Can you at least show some work and your own thoughts on these problems? It'd be easier for us to point you in the direction than starting right from scratch.

1. I dunno if i have to: find a formula for P, or have to do it with h. Anyways, i have no clue how to do it.

2. I got a guess, it's: x/((n+1)x+1). i got it by observing f1,f2...fn. it seems that the coefficient of x on the denominator only increments by 1. can someone verify it?

3. i already solved it. and im a hunded percent sure of my answer!

4. i thought about graphing the equation by replacing c to y, but from the graph, dont know how to extract the answer from it.

5. have zero clue on how to do it.

6. the problem here is that i can't think of an integral of (sin t)/t..

7.a. is there a property where f(a-b)=f(a)-f(b)?

b. if i solve a., this would be easy.

4. ## I love this problem.

Originally Posted by superben
Can you help me solve my problem set please? I already solved no 3 and am working on the others. but as of now, finishing it is far and deadline's on monday. Really need to finish this all but can't do it without help from you guys. please help me. thanks!

or see attachments below

Here is #7

$\int_0^af(a-x)dx$ let u= a-x $du=-dx$

so

$\int_0^af(a-x)dx=\int_a^0f(u)(-du)=\int_0^af(u)du$

Now for the fun Part

$\int\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}dx=\int\f rac{\sin^n(\pi/2-x)}{\cos^n(\pi/2-x)+\sin^n(\pi/2-x)}dx$

but

$\sin(\pi/2-x)=\cos(x)$ and $\cos(\pi/2-x)=sin(x)$

$\int\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}dx=\int\f rac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}dx= L$ Then

$\int\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}dx+\int\f rac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}dx =2L$

summing up the left side and reducing we get...

$\int_0^{\pi/2}dx=2L$ or

$\frac{\pi}{2}=2L \iff \frac{\pi}{4}=L$

YEAH!!!

5. 5. $\frac{\left(x^{2} + 1\right)\left(y^{2} + 1\right)\left(z^{2} + 1\right)}{xyz} = \left(\frac{x^{2} + 1}{x}\right) \cdot \left(\frac{y^{2} + 1}{y}\right) \cdot \left(\frac{z^{2} + 1}{z}\right)$

My hint to you: What's the minimum value that $\frac{a^{2} + 1}{a}$ can have for a > 0? Use your first derivative test.

@TheEmptySet: Haha I loved how enthusiastic you made the problem seem. "YEAH!!!"

6. Originally Posted by TheEmptySet
Here is #7

$\int_0^af(a-x)dx$ let u= a-x $du=-dx$

so

$\int_0^af(a-x)dx=\int_a^0f(u)(-du)=\int_0^af(u)du$

Now for the fun Part

$\int\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}dx=\int\f rac{\sin^n(\pi/2-x)}{\cos^n(\pi/2-x)+\sin^n(\pi/2-x)}dx$

but

$\sin(\pi/2-x)=\cos(x)$ and $\cos(\pi/2-x)=sin(x)$

$\int\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}dx=\int\f rac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}dx= L$ Then

$\int\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}dx+\int\f rac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}dx =2L$

summing up the left side and reducing we get...

$\int_0^{\pi/2}dx=2L$ or

$\frac{\pi}{2}=2L \iff \frac{\pi}{4}=L$

YEAH!!!
I'm not really sure if you proved 7.a. clearly. you just equated it an integral of f of u, which is not whta the problem asked, it should be an f of x. please clear this out, but thanks for the help as well!

7. Originally Posted by o_O
5. $\frac{\left(x^{2} + 1\right)\left(y^{2} + 1\right)\left(z^{2} + 1\right)}{xyz} = \left(\frac{x^{2} + 1}{x}\right) \cdot \left(\frac{y^{2} + 1}{y}\right) \cdot \left(\frac{z^{2} + 1}{z}\right)$

My hint to you: What's the minimum value that $\frac{a^{2} + 1}{a}$ can have for a > 0? Use your first derivative test.

@TheEmptySet: Haha I loved how enthusiastic you made the problem seem. "YEAH!!!"
so f'(x) = (x^2-1)/x^2 ayt?

0 = (x^2-1)/x^2

x = 1, because x should always be >0.

hmm.. so f(1) = ((1)^2 + 1)/(1)..

and that would be a miraculous 2!! wow! that was not a hint, that was an answer! thanks a lot!!!! Hope you'll help me with the others too!

8. 6 You don't have to find the indefinite integral of sinx/x at all (in fact, you won't be able to but we won't get into that). For any integrable function, what would $\int_{a}^{a}f(x)dx$ evaluate to?

9. Originally Posted by o_O
6 You don't have to find the indefinite integral of sinx/x at all (in fact, you won't be able to but we won't get into that). For any integrable function, what would $\int_{a}^{a}f(x)dx$ evaluate to?
$\int_{a}^{a}f(x)dx$? zero right?

will that mean i can now use L'Hospital's rule? because if $\int_{a}^{a}f(x)dx$ is zero, then x multiplied by it is zero too? (with the denominator, it's obviously zero) so all i have to do is... use L'Hospital's rule! Yes, i think that what you're thinking too! A BIG tHANKS! Thank you very much! (4 to go, still not sure with TheEmptySet's answer)

is the answer for no.6 sin 3?? please someone confirm it thanks

10. Looks good to me. Using l'hopital leads to using FTC which you seem to have gotten so that's good.

Sorry if I'm just doing random questions but I don't have time to go through everything in one sitting and just picking random ones I see.

2. That looks good to me as well. I'm not sure if you have to prove it or not but you could probably do it with induction.

1. Let the triangle have sides x, y, and h. So:
$P = x + y + h \quad \mbox{Perimeter = sum of its sides}$
$25 = \frac{1}{2}xy \quad \Rightarrow 50 = xy$

Now, by the Pythagorean theorem: $h^{2} = x^{2} + y^{2}$. Since we're dealing with $h^{2}$, let's square P as well:
$P^{2} = (x + y + h)^{2} = x^{2} + y^{2} + h^{2} + 2xy + 2hx + 2hy$

Now you should be able to group the right side into terms of P and h. AFter that, solve for h.

11. Originally Posted by o_O
Looks good to me. Using l'hopital leads to using FTC which you seem to have gotten so that's good.

Sorry if I'm just doing random questions but I don't have time to go through everything in one sitting and just picking random ones I see.

2. That looks good to me as well. I'm not sure if you have to prove it or not but you could probably do it with induction.

1. Let the triangle have sides x, y, and h. So:
$P = x + y + h \quad \mbox{Perimeter = sum of its sides}$
$25 = \frac{1}{2}xy \quad \Rightarrow 50 = xy$

Now, by the Pythagorean theorem: $h^{2} = x^{2} + y^{2}$. Since we're dealing with $h^{2}$, let's square P as well:
$P^{2} = (x + y + h)^{2} = x^{2} + y^{2} + h^{2} + 2xy + 2hx + 2hy$

Now you should be able to group the right side into terms of P and h. AFter that, solve for h.
2. Yes, i think mathematical induction is best suited for this, but don't know how for now.
1. i got this based from what you've said: h = (P^2 - 100)/2P and it seems to be right

got any clue for nos. 4 and 7?

12. ## Probset Stats

Already solved numbers 1, 3, 5, and 6.
Numbers not yet solved are 2, 4, 7.a, and 7.b..

2: the formula is x/((n+1)x+1) but i still don't know how to prove it with induction yet.
4: graphing it seems to solve the problem, but still, can't extract the answer there
7.a, 7.b: EmptySet's answer is still not clear. can someone explain it?

THANKS!

13. Originally Posted by superben
I'm not really sure if you proved 7.a. clearly. you just equated it an integral of f of u, which is not whta the problem asked, it should be an f of x. please clear this out, but thanks for the help as well!

$
\int_0^af(a-x)dx=\int_a^0f(u)(-du)=\int_0^af(u)du
$

If it makes you happy use the substituion u=x for the last integral

Remember $\int_0^af(t)dt=\int_0^af(x)dx$

$u=x \mbox{ and } du =dx$

$
\int_0^af(a-x)dx=\int_a^0f(u)(-du)=\int_0^af(u)du=\int_0^af(x)dx
$

14. 4. If you consider $c \leq 0$, what can you deduce from the graphs? If you consider c > 0, then the parabola must touch ln(x) at one point and curve away. So, there is a line that passes through this point that acts as a tangent to both curves. i.e. $\frac{d}{dx} \ln x = \frac{d}{dx} cx^{2}$.

2. $P(n): f_{n+1}(x) = \left( f_{0} \circ f_{n}\right)(x) = \frac{x}{(n+2)x + 1}\quad \forall n \geq 0$

$P(0) : f_{1}(x) = \left( f_{0} \circ f_{0}\right)(x) = \frac{x}{2x+1}$

$\mbox{Assume } P(k) : f_{k+1}(x) = \left( f_{0} \circ f_{k}\right)(x) = \frac{x}{(k+2)x + 1}$

We must prove that:
$P(k+1) : f_{k+2} = \left( f_{0} \circ f_{k+1}\right)(x) = \frac{x}{(k+3)x + 1}$

This should be easy. Just simple algebraic manipulation

15. i really can't solve no.4
and no. 2 to..
head seems to be not working anymore.

wahhh..

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