Question

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Answer 1

Răspuns:

a)

$\displaystyle\int_0^1\left(x^4+1\right)f(x)dx=\int_0^1(x^4+1+2x)dx=\left.\frac{x^5}{5}\right|_0^1+\left.x\right|_0^1+\left.x^2\right|_0^1=\frac{1}{5}+1+1=\frac{11}{5}$

b) Dacă F are asimptotă oblică spre $\infty$ atunci $\displaystyle\lim_{x\to\infty}F(x)=\infty$ sau $\displaystyle\lim_{x\to\infty}F(x)=-\infty$

Atunci

$m=\displaystyle\lim_{x\to\infty}\frac{F(x)}{x}\overset{\frac{\infty}{\infty}}{=}\lim_{x\to\infty}\frac{F'(x)}{1}=\lim_{x\to\infty}f(x)=1$

c)

$G(x)=\displaystyle\int\left(1+\frac{2x}{x^4+1}\right)dx=x+\int\frac{2x}{(x^2)^2+1}dx=x+\arctan(x^2)+\mathcal{C}$

Din $G(0)=0\Rightarrow \mathcal{C}=0\Rightarrow G(x)=x+\arctan{(x^2)}$

Atunci

$\displaystyle\int_0^1xG(x)dx=\int_0^1\left(\frac{x^2}{2}\right)'G(x)dx=\left.\frac{x^2}{2}G(x)\right|_0^1-\int_0^1\frac{x^2}{2}f(x)dx=\frac{G(1)}{2}-\left.\frac{1}{4}\ln(x^4+1)\right|_0^1=\frac{1}{3}+\frac{\pi}{8}-\frac{1}{4}\ln 2$

Explicație pas cu pas: