denne var litt vrien
Grenseverdiene er 4 til 1.
over telleren skal være, e opphøyd i[tex] \sqrt{x}[/tex]
[tex]\int \frac{e^\sqrt{x}}{2\sqrt{x}} dx[/tex]
integral2
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]\int_1^4 \frac{e^{\sqrt{x}}}{2\sqrt{x}}\rm{d}x[/tex]
[tex]u = \sqrt{x} \ , \ u^, = \frac{1}{2\sqrt{x}} \ , \ \frac{\rm{d}u}{\rm{d}x} = \frac{1}{2u} \ , \ \rm{d}x = 2u\rm{d}u[/tex]
[tex]\int \frac{e^{u}}{\cancel{2u}} \ \cdot \ \cancel{2u}\rm{d}u = \int e^u\rm{d}u = e^u + C[/tex]
[tex]\int_1^4 \frac{e^{\sqrt{x}}}{2\sqrt{x}}\rm{d}x = [e^{\sqrt{x}}]_1^4[/tex]
[tex]u = \sqrt{x} \ , \ u^, = \frac{1}{2\sqrt{x}} \ , \ \frac{\rm{d}u}{\rm{d}x} = \frac{1}{2u} \ , \ \rm{d}x = 2u\rm{d}u[/tex]
[tex]\int \frac{e^{u}}{\cancel{2u}} \ \cdot \ \cancel{2u}\rm{d}u = \int e^u\rm{d}u = e^u + C[/tex]
[tex]\int_1^4 \frac{e^{\sqrt{x}}}{2\sqrt{x}}\rm{d}x = [e^{\sqrt{x}}]_1^4[/tex]
