Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x^{2}-4$.

$5 \mathbf{p}$ a) Arătați că [tex]$\int_{0}^{3}(f(x)+4) d x=9$[tex].

$5 p$ b) Calculați $\int_{0}^{1} \frac{1}{f(x)+5} d x$.

$5 p$ c) Determinaţi numărul real $a, a\ \textgreater \ 0$, pentru care $\int_{\frac{1}{a}}^{a} f\left(\frac{1}{x}\right) d x=-8$.

Answer 1

Răspuns:

a) $\displaystyle\int_0^3(f(x)+4)dx=\int_0^3x^2dx=\left.\frac{x^3}{3}\right|_0^3=9$

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

c) $\displaystyle\int_{\frac{1}{a}}^af\left(\frac{1}{x}\right)dx=\int_{\frac{1}{a}}^a\left(\frac{1}{x^2}-4\right)dx=\left. -\frac{1}{x}\right|_{\frac{1}{a}}^a-\left. 4x\right|_{\frac{1}{a}}^a=\frac{-3a^2+3}{a}=-8$

Rezultă $a=3$.

Explicație pas cu pas: