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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. –  Davide Giraudo Aug 18 '11 at 12:54
    
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. –  Did Aug 18 '11 at 12:58
    
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. –  newbie Aug 18 '11 at 13:02
    
@DavideGiraudo and newbie: Thank you. I will try that when I get back home. –  entrance_exam Aug 18 '11 at 13:09
    
@DidierPiau I wanted to show my result (or absent of it) to ask for verification. (how could I improve?) –  entrance_exam Aug 18 '11 at 13:09
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1 Answer

up vote 3 down vote accepted

$$\displaystyle f(x) = \frac1{x^2+2x+4} = \frac1{(x+1)^2 + 3}$$ 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)? –  entrance_exam Aug 20 '11 at 8:04
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