Question

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

$5 p$ a) Arătați că $\int_{0}^{1} f(x) d x=\frac{5}{6}$.
$5 \mathbf{b} \mid$ b) Arătați că $\int_{1}^{e} \frac{f(x)}{x} \ln x d x=\frac{e^{2}-7}{4}$
5p c) Determinați numerele reale $a, a\ \textgreater \ 1$ pentru care $\int_{1}^{a} f(x) e^{x} d x=e^{a}-3 e$

Answer 1

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

a)

$\int\limits^1_0 {x^2-3x+2} \, dx =\int\limits^1_0 x^2\ dx-\int\limits^1_0 3x\ dx+\int\limits^1_0 2\ dx=\frac{x^3}{3}|_0^1-\frac{3x^2}{2}|_0^1+2x|_0^1=\frac{1}{3} -\frac{3}{2}+2=\frac{5}{6}$

(vezi tabel de integrale)

b)

$\int\limits^e_1 {\frac{x^2-3x+2}{x}lnx } \, dx$

O desfacem in doua integrale

$\int\limits^e_1 {\frac{x^2-3x}{x}lnx } \, dx+\int\limits^e_1 {\frac{2}{x}lnx } \, dx$

Le luam pe rand

$\int\limits^e_1 {\frac{x^2-3x}{x}lnx } \, dx=\int\limits^e_1 {\frac{x(x-3)}{x}lnx } \, dx=\int\limits^e_1 (x-3)lnx\ dx$

O integram prin parti

$f=lnx\ \ \ \ \ f'=\frac{1}{x} \\\\g'=x-3\ \ \ \ \ g=\frac{x^2}{2}-3x=\frac{x(x-6)}{2}$

$\int\limits^e_1 (x-3)lnx\ dx=\frac{x(x-6)}{2}lnx\ |_1^e -\frac{1}{2} \int\limits^e_1x-6\ dx=\frac{x(x-6)}{2}lnx\ |_1^e -\frac{1}{2}(\frac{x^2}{2}-6x)|_1^e=\frac{e(e-6)}{2}-\frac{1}{2}(\frac{e^2}{2}-6e-\frac{1}{2}+6)=\\\\ =\frac{2e^2-12e-e^2+12e+1-12}{4}=\frac{e^2-11}{4}$

Luam cea de-a doua integrala

$\int\limits^e_1 {\frac{2}{x}lnx } \, dx=\int\limits^e_1 {2(lnx)'lnx } \, dx=ln^2x|_1^e=ln^2e-ln^21=1$

Integrala noastra din cerinta ne va da:

$\frac{e^2-11}{4}+1=\frac{e^2-7}{4}$

c)

$\int\limits^a_1 {(x^2-3x+2)e^x} \, dx$

Desfacem in 3 integrale si le calculam separat:

$\int\limits^a_1 {x^2e^x} \, dx$

$f=x^2\ \ \ \ \ f'=2x\\\\g'=e^x\ \ \ \ g=e^x$

$\int\limits^a_1 {x^2e^x} \, dx=x^2e^x|_1^a-\int\limits^a_1 {2xe^x} \, dx$

$\int\limits^a_1 {(x^2-3x+2)e^x} \, dx=x^2e^x|_1^a-\int\limits^a_1 {2xe^x} \, dx-\int\limits^a_1 {3xe^x} \, dx+\int\limits^a_1 {2e^x} \, dx=a^2e^a-e-\int\limits^a_1 {5xe^x} \, dx+2e^a-2e$

$\int\limits^a_1 {5xe^x} \, dx=5xe^x|_1^a-\int\limits^a_1 {5e^x} \, dx =5ae^a-5e-5e^a+5e=5ae^a-5e^a$

$\int\limits^a_1 {(x^2-3x+2)e^x} \, dx=a^2e^a-e-(5ae^a-5e^a)+2e^a-2e=e^a(a^2-5a+7)-3e$

$e^a(a^2-5a+7)-3e=e^a-3e\\\\a^2-5a+7=1\\\\a^2-5a+6=0$

Δ=25-24=1

a=3

a=2

Un exercitiu cu integrale gasesti aici: https://brainly.ro/tema/956588

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