Question

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

$5 \mathbf{p}$ a) Arătați că $\int_{0}^{2} f(x) e^{-x} d x=4$. 5p b) Calculați $\int_{1}^{e} \frac{1}{x^{2}} f(\ln x) d x$. $5 p$ c) Arătați că $\int_{0}^{1} f(x) F(x) d x=2(e-3)^{2}$, unde $F: \mathbb{R} \rightarrow \mathbb{R}$ este primitiva funcţiei [tex]$f$[tex] pentru care
$F(0)=0 .$

Answer 1

$f(x)=x^{3} e^{x}$

a)

Ne folosim de tabelul de integrale (vezi atasament)

$\int\limits^2_0 {x^3e^xe^{-x}} \, dx =\int\limits^2_0 x^3\ dx=\frac{x^4}{4} |_0^2=4-0=4$

b)

$\int\limits^e_1 {\frac{ln^3(x)e^{lnx}}{x^2} } \, dx \\\\e^{lnx}=x\\\\\int\limits^e_1 {\frac{ln^3(x)}{x} } \, dx\\\\(lnx)'=\frac{1}{x} \\\\\int\limits^e_1 {ln^3(x)(lnx)'} \, dx=\frac{1}{4} \int\limits^e_1 {4ln^3(x)(lnx)'} \, dx=\frac{1}{4}ln^4(x)|_1^e=\frac{1}{4} -0=\frac{1}{4}$

c)

F(x) este primitiva functiei f, adica F'(x)=f(x)

$\int\limits^1_0 {f(x)F(x)} \, dx =\int\limits^1_0 {F'(x)F(x)} \, dx =\frac{1}{2}F^2(x)|_0^1=\frac{1}{2}(F^2(1)-F^2(0))=\frac{1}{2}F^2(1)$

Vom calcula F(x), apoi F(1)

$F(x)=\int\limits {x^3e^x} \, dx$

f=x³           f'=3x²

g'=eˣ         g=eˣ

$F(x)=x^3e^x-\int\limits3x^2e^x {x} \, dx$

Luam separat integrala $3\int\limits x^2e^x {x} \, dx$

f=x²           f'=2x

g'=eˣ         g=eˣ

$\int\limits3x^2e^x {x} \, dx =3(x^2e^x-\int2xe^x\ dx)$

Luam separat integrala $2\int xe^x\ dx$

f=x         f'=1

g'=eˣ      g=eˣ

$\int2xe^x\ dx=2(xe^x-\int e^x\ dx)=2xe^x-2e^x$

Ne intoarcem mai sus si inlocuim ce am obtinut:

$\int\limits3x^2e^x {x} \, dx =3(x^2e^x-2xe^x+2e^x)=3x^2e^x-6xe^x+6e^x$

$F(x)=x^3e^x-3x^2e^x+6xe^x-6e^x=e^x(x^3-3x^2+6x-6)+c\\\\\ F(0)=0\\\\\ F(0)=1(0-0+0-6)+c=-6+c=0\\\\\ c=6\\\\\ F(x)=e^x(x^3-3x^2+6x-6)+6\\\\\\F(1)=e(1-3+6-6)+6=-2e+6=-2(e-3)\\\\F^2(1)=4(e-3)^2$

Deci $\int\limits^1_0 {f(x)F(x)} \, dx =\int\limits^1_0 {F'(x)F(x)} \, dx =\frac{1}{2}F^2(x)|_0^1=\frac{1}{2}(F^2(1)-F^2(0))=\frac{1}{2}F^2(1)=\frac{1}{2}4(e-3)^2=2(e-3)^2$

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/2498443

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