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I worked out a solution for a problem I am studying for an entrance exam, but I am not sure if it is correct. I would appreciate if some knowledgeable person could help me out.

Given

$$ \int_0^x t^{2}f(t)dt + 2 \int_0^x tf(t)dt = x-4 \int_0^x f(t)dt $$

  1. Obtain $f(x)$.
  2. Obtain the maximum Value of $f(x)$.
  3. Calculate the improper integral $\displaystyle \int_0^{+\infty} f(t)dt $.

I differentiated both sides (using the Fundamental Theorem of Calculus) and solved for $f(x)$. Came up with this: $$ x^{2}f(x) + 2xf(x) = 1 - 4 f(x) $$ $$ f(x) = 1/(x^{2}+2x+4) $$

I tried integrating it but it is complicated so I used the formula I got after differentiating

$$ x^{2}f(x)+2xf(x) = 1 - 4f(x) $$

Setting $x$ to $1$ solved the equation. (So I think it is right, but not sure)

For 2. my answer is $\frac 14$ because $f(0)$ is $\frac 14$. If $x$ increases it will tend to $0$.

For 3. I don't have a solution because I did not manage to integrate the function yet.

Any help or advice is appreciated. How can I integrate $f(x)$ to calculate the improper integral. Is $f(x)$ right at all?

Best regards

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    You can write $\frac 1{x^2+2x+4} =\frac 1{(x+1)^2+3}$: it will help you to find the maximum value, and to integrate this function.2011-08-18
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    Like Davide said. // Up to just before *I tried integrating it but...*, the post is perfect, after that point I do not see how it relates to the questions asked.2011-08-18
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    and hint for 3: use $u = x+1$ and $v = \frac{u}{\sqrt{3}}$. the result of the integral I found is $\frac{\pi}{3\sqrt{3}}$. I hope it is correct.2011-08-18
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    @DavideGiraudo and newbie: Thank you. I will try that when I get back home.2011-08-18
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    @DidierPiau I wanted to show my result (or absent of it) to ask for verification. (how could I improve?)2011-08-18
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    @entrance: Yes, and you are absolutely right to do so! My aim was to localize the place of the problems in your post (and the places that are OK). So... now that you have an expression for $f$, is question 2. allright? Davide's hint should get you the value of $x$ where $f(x)$ is maximum... Please continue explaining the obstacles you meet, if any.2011-08-18
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    @DidierPiau Thank you. I was able to understand the solution below and redo it. The hardest part was finding out why the lower limit of the integral should be PI/6. I managed to get this value by using the unit circle. Is there any easier way than memorizing the unit circle?2011-08-20
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    @entrance: you mean, do you have to know tan(pi/6)? My answer is **yes, absolutely**. One should at least know the values of sine and cosine of 0, pi/6, pi/4, pi/3 and pi/2 and how to (re)check and/or extend them to the lines tangent and cotangent and to other values, replacing theta by -theta or pi+theta or pi-theta or pi/2+theta or pi/2-theta.2011-08-20
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    @Didier Thanks! That's exactly what I wanted to know. I will study them.2011-08-20

1 Answers 1

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$$\displaystyle f(x) = \frac1{x^2+2x+4} = \frac1{(x+1)^2 + 3}$$ Note that $(x+1)^2 \geq 0$ with the equality holding at $x=-1$.

This gives us $(x+1)^2 + 3 \geq 3$ and hence $\displaystyle \frac1{(x+1)^2 + 3} \leq \frac1{3}$ with the equality holding at $x=-1$.

Hence, the maximum value of $f(x) = \frac1{3}$ which is obtained at $x=-1$.

To integrate $f(x)$ from $0$ to $\infty$, let $(x+1) = \sqrt{3} \tan (\theta)$.

Note that $x = 0 \implies \theta = \pi/6$ and $x = \infty \implies \theta = \pi/2$

$$I = \int_{0}^{\infty} \frac1{(x+1)^2 + 3} dx = \int_{\pi/6}^{\pi/2} \frac{\sqrt{3} \sec^{2}(\theta)}{3 \tan^2(\theta) + 3} d \theta = \frac1{\sqrt{3}} \int_{\pi/6}^{\pi/2} \frac{\sec^{2}(\theta)}{\tan^2(\theta) + 1} d \theta$$

$$I = \frac1{\sqrt{3}} \int_{\pi/6}^{\pi/2} \frac{\sec^{2}(\theta)}{\sec^2(\theta)} d \theta = \frac1{\sqrt{3}} \int_{\pi/6}^{\pi/2} d \theta = \frac1{\sqrt{3}} \left( \frac{\pi}{2} - \frac{\pi}{6}\right) = \frac{\pi}{3 \sqrt{3}}$$

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    Great, thank you! I was finally able to understand it. How exactly did you calculate the lower limit of the integral? PI/6. Is there a way to easily know when tan will be 1/sqrt(3)?2011-08-20