Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=3 x^{3}+4 x^{2}$.

5p a) Arătaţi că $\int_{0}^{2}\left(f(x)-4 x^{2}\right) d x=12$.

$5 \mathbf{p}$ b) Determinaţi primitiva $F$ a funcţiei $f$ pentru care $F(0)=2020$.

$5 p$ c) Determinaţi numărul real $m, m\ \textgreater \ 1$, ştiind că $\int_{1}^{m} \frac{f(x)}{x^{2}} d x=\frac{17}{2}$.

Answer 1

$f(x)=3 x^{3}+4 x^{2}$

a)

Vezi tabelul de integrale din atasament

$\int\limits^2_0 {3x^3} \, dx =\frac{3x^4}{4}|_0^2=\frac{3\cdot 16}{4} -0=12$

b)

F'(x)=f(x)

F(0)=2020

$F(x)=\int\limits {f(x)} \, dx =\frac{3x^4}{4}+\frac{4x^3}{3}+C\\\\ F(x)=\frac{3x^4}{4}+\frac{4x^3}{3} +C\\\\F(0)=0+0+C=2020\\\\\ C=2020\\\ F(x)=\frac{3x^4}{4}+\frac{4x^3}{3} +2020$

c)

$\int\limits^m_1 {\frac{3x^3+4x^2}{x^2} } \, dx= \int\limits^m_13x\ dx+\int\limits^m_14 \ dx=\frac{3x^2}{2}|_1^m+4x|_1^m=\frac{3m^2}{2}-\frac{3}{2}+4m-4 \\\\ \frac{3m^2}{2}-\frac{3}{2}+4m-4 =\frac{17}{2}\\\\ 3m^2-3+8m-8=17\\\\ 3m^2+8m-28=0\\\\\Delta=64+336=400\\\\m_1=\frac{-8+20}{6} =2\\\\m_2=\frac{-8-20}{6} < 1\ NU$

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9928472

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