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Question Number 183450 by cortano1 last updated on 26/Dec/22

$$\underset{0}{\stackrel{x^2}{\int }}f(2t^3-t^2-6)dt=\ln x$$$$thenf\left(106\right)=?$$

Answered by mahdipoor last updated on 26/Dec/22

$$\frac{d}{dx}L=\frac{d}{dx}R\Rightarrow f(2x^6-x^2-6)\times 2x=\frac{1}{x}$$$$x=2\Rightarrow f\left(106\right)=\frac{1}{8}$$$$-------$$$$\frac{d}{dx}{\int }_0^{g\left(x\right)}f\left(h\left(t\right)\right)dt=\frac{{\int }_{g\left(x\right)}^{g(x+\mathrm{\Delta }x)}f\left(g\left(t\right)\right)dt}{\mathrm{\Delta }x}\mathrm{\Delta }x\to 0$$$$\Rightarrow (get\int f\left(h\left(t\right)\right)dt=k\left(t\right))\Rightarrow$$$$=\frac{k\left(g(x+\mathrm{\Delta }x)\right)-k\left(g\left(x\right)\right)}{\mathrm{\Delta }x}=k^{{}^{\prime }}\left(g\left(x\right)\right)g^{{}^{\prime }}\left(x\right)$$$$=f\left(h\left(g\left(x\right)\right)\right)g^{{}^{\prime }}\left(x\right)$$$$\Rightarrow h\left(t\right)=2t^3-t^2-6g\left(x\right)=x^2\Rightarrow$$$$f(2x^6-x^4-6)\times 2x$$$$$$

Answered by mr W last updated on 26/Dec/22

$$sayf(2t^3-t^2-6)=g\left(t\right)$$$${\int }_0^{x^2}g\left(t\right)dt=\ln x$$$$\frac{d}{dx}\left[{\int }_0^{x^2}g\left(t\right)dt\right]=\frac{d}{dx}\left(\ln x\right)$$$$g\left(x^2\right)\left(2x\right)=\frac{1}{x}$$$$\Rightarrow g\left(x^2\right)=\frac{1}{2x^2}$$$$\Rightarrow g\left(t\right)=\frac{1}{2t}$$$$\Rightarrow f(2t^3-t^2-6)=\frac{1}{2t}$$$$set2t^3-t^2-6=106$$$$2t^3-t^2-112=0$$$$(t-4)(2t^2+7t+28)=0$$$$\Rightarrow t=4$$$$\Rightarrow f\left(106\right)=\frac{1}{2\times 4}=\frac{1}{8}✓$$

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