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- Feb 14, 2012

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$\displaystyle \int_1^2 f^2(t^3)dt + 2\int_1^2 f(t^3)dt=\dfrac{2}{3}\int_1^8 f(t)dt-\int_1^2 (t^2-1)^2 dt$

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- Thread starter
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- #1

- Feb 14, 2012

- 3,802

$\displaystyle \int_1^2 f^2(t^3)dt + 2\int_1^2 f(t^3)dt=\dfrac{2}{3}\int_1^8 f(t)dt-\int_1^2 (t^2-1)^2 dt$

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- Feb 14, 2012

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$\displaystyle \dfrac{2}{3} \int_1^8 f(t)dt=2\int_1^2 u^2f(u^3)du=2\int_1^2 t^2f(t^3)du$

Hence, by the assumptions,

$\displaystyle \int_1^2 [f^2(t^3)+(t^2-1)^2+2f(t^3)-2t^2f(t^3)] dt=0$

Since $f^2(t^3)+(t^2-1)^2+2f(t^3)-2t^2f(t^3)=[f(t^3)]^2+(1-t^2)^2+2(1-t^2)f(t^3)=[f(t^3)+1-t^2]^2\ge 0$, we get

$\displaystyle \int_1^2 [f(t^3)+1-t^2]^2 dt=0$

The continuity of $f$ implies that $f(t^3)=t^2-1,\,1\le t \le 2$ thus $f(x)=x^{\tiny\dfrac{2}{3}}-1,\,1 \le x \le 8$.