Question

f(x)=e^x+x^2+2x şi F(x)=e^x+x^3/3+x^2+1

a) să se arate că funcţia F este o primitivă a functiei F.

b) să se calculeze integrală (0.1) f(x)dx

Answer 1

a) Trebuie f . F este o primitiva pentru f daca f este derivabila  ⇒  F'(x) = f(x) 

Astfel calculam derivata functiei F.

[tex] F^{'}(x) = ( e^{x} + \frac{x^{3} }{3} + x^{2} + 1 ) ' = [tex] (e^{x})' + (\frac{x^{3} }{3})' + (x^{2})+ (1)' = e^{x} + x^{2} +2x +0 = e^{x} + x^{2} +2x [/tex].  

Deci F'(x) = f(x)

b) $\int\limits^0_1 {f(x)} \, dx = \int\limits^0_1 { (e^{x} + x^{2}+2)x} \, dx$ = $\int\limits^0_1 { e^{x}} \, dx + \int\limits^0_1 { x^{2} } \, dx + \int\limits^0_1 {2x} \, dx =$
continuarea in imagine 

Answer 2

$\displaystyle \mathtt{f(x)=e^x+x^2+2x,~F(x)=e^x+ \frac{x^3}{3} +x^2+1}\\ \\ \mathtt{a)~~~F'(x)=\left(e^x+ \frac{x^3}{3} +x^2+1\right)'=\left(e^x\right)'+\left( \frac{x^3}{3} \right)'+\left(x^2\right)'+1'=}\\ \\ \mathtt{=e^x+ \frac{1}{3} \cdot \left(x^3\right)'+2x+0=e^x+ \frac{1}{3} \cdot 3x^2+2x=e^x+x^2+2x=f(x)\Rightarrow }\\ \\ \mathtt{\Rightarrow F~este~o~primitiv\u{a}~a~func\c{t}iei ~f.}$

$\displaystyle \mathtt{b)~~~\int\limits_0^1f(x)dx=\int\limits_0^1\left(e^x+x^2+2x\right)dx=\int\limits_0^1e^xdx+\int\limits_0^1x^2dx+2\int\limits_0^1xdx=}\\ \\ \mathtt{=e^x\Bigg|_0^1+ \frac{x^3}{3}\Bigg|_0^1+2\cdot\frac{x^2}{2} \Bigg|_0^1=\left(e^1-e^0\right)+\left( \frac{1^3}{3}- \frac{0^3}{3} \right)+2\left( \frac{1^2}{2}- \frac{0^2}{2}\right)=}\\ \\ \mathtt{=e-1+ \frac{1}{3}+2 \cdot \frac{1}{2}=e-1+ \frac{1}{3}+1= \frac{3e-3+1+3}{3}= \frac{3e+1}{3} }$