3
$\begingroup$

Would you please tell me whether there is any wrong on this problem? given that $g$ is continuous on $[0,\infty)\rightarrow \mathbb{R}$ satisfying $\int_{0}^{x^2(1+x)}g(t)dt=x \forall x\in [0,\infty)$ then I need to find what is $g(2)$?

  • 0
    Not quite --- see the comments below for an answer.2012-06-03

2 Answers 2

3

Let $G(x) = \int _{0} ^{x} g(t) dt$. Since $g(t)$ is continuous, we can deduce that $G(x)$ exists and will be differentiable for $x \ge 0$.

Then, by your condition, $G(x^2(1+x))=x$.

Differentiating both sides with respect to $x$, we get $(2x(1+x)+x^2) \times g(x^2(1+x))=1$.

Simplifying, we get $g(x^2(1+x))= \frac{1} {3x^2+2x}$

Now, we set $(x^2(1+x))=2$. This is true when $x=1$ and has two imaginary roots at $x=i-1$ and $x=-i-1$.

We are looking for real solutions, so $x=1$ and $g(2)=\frac {1} {5}$

2

Note that $ \frac{d}{dx}\int_0^{x^2(1+x)}g(t)\,dt=g(x^2(1+x))(2x+3x^2)=\frac{d}{dx}x=1. $ So now solve for $g(2)$.