Răspuns:
I=∫∫Dxydxdy
y=x²
ay-bx=0
Faci inlocuirile
5y-4x=0
Determini limitele de integrare
5*x²-4x=0
x(5x-4)=0
x1=0
x2=5/4
Observi conf regulii semnelr pt functia de gradul 2 ca expresia e negativa (esti intre radacini)=> dreapa 5y-4x=0 este deasupra parabolei y=x²
I=∫∫xydx dy=
∫([tex]\int\limits^\frac{5x}{4} _ {x^2} \,xy dy))dx[/tex]
[tex]\int\limits^\frac{5x}{4} _ {x^2} xy\, dx =[/tex]
[tex]x\int\limits^\frac{5x}{4} _{x^2} \,y dx =[/tex]
x[tex]\frac{y^2}{2}[/tex]║⁵ˣ/4ₓ²=
x*[tex](\frac{5x}{4} )^2-x^2)[/tex]=
x([tex]\frac{25x^2}{16} -x^2)[/tex]
[tex]x\frac{9x^2}{16}[/tex]=
[tex]\frac{9x^3}{16} =[/tex]
I=[tex]\int\limits^\frac{5}{4} _ {0} \,\frac{9x^3}{16} dx =[/tex]
[tex]\frac{9}{16} \int\limits^\frac{5}{4} _ {0} \,x^3 dx[/tex]
[tex]\frac{9}{16} \frac{x^4}{4}[/tex]║₀⁵/4=
[tex]\frac{9}{16} *[(\frac{5}{4} )^4-0][/tex]
[tex]\frac{9}{4^2} *\frac{5^4}{4^4}[/tex]
[tex]\frac{9*625}{4^6}[/tex]
Explicație pas cu pas:
