Let $f:[0,\infty)\rightarrow \mathbb{R}$ be continuous function satisfying $\int_{0}^{f(x)}t^2dt=x^3(1+x)^2$ then what is $f(2)$, what I did: I put $x=2$ both side and got $f(2)=6$, can anyone check it for me?
A question on definite integral to find a value, just to confirm myself
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real-analysis
definite-integrals
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0linked to [this](http://math.stackexchange.com/questions/153409/a-question-on-definite-integral-to-find-a-value) – 2012-06-03
2 Answers
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If you integrate it out, you will get $$\dfrac{f(x)^3}{3} = x^3(1+x)^2$$ Hence, $$f(x)^3 = 3x^3(1+x)^2$$ Setting $x = 2$, gives us $$f(2)^3 = 3 \times 2^3 \times 3^2 = 6^3.$$ Since $f(x) \in \mathbb{R}$, we get that $f(2) = 6$.
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$$\int_{0}^{f(x)}t^2dt=\frac{{f(x)}^3}{3}=x^3(1+x)^2$$
$$\Downarrow$$
$${f(x)}^3=3x^3(1+x)^2$$ $$\Downarrow$$
$${f(2)}^3=3 \cdot2^3(1+2)^2=3^3 \cdot 2^3$$ $$\Downarrow$$
$$f(2)=3 \cdot 2 =6$$