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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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    @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