determinar a função f tal que integral de f(x)dx=x^2+1/2cos(2x)+c
Soluções para a tarefa
Respondido por
10
É só derivar. Acompanhe:
![\displaystyle\mathsf{ y= x^2+\frac{1}{2} \cos 2x + C } \\ \\ \\ \mathsf{y' = 2x + \frac{1}{2} \cdot (-\sin 2x \cdot 2)} \\ \\ \\ \mathsf{y' = 2x - \sin 2x} \displaystyle\mathsf{ y= x^2+\frac{1}{2} \cos 2x + C } \\ \\ \\ \mathsf{y' = 2x + \frac{1}{2} \cdot (-\sin 2x \cdot 2)} \\ \\ \\ \mathsf{y' = 2x - \sin 2x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathsf%7B++y%3D+x%5E2%2B%5Cfrac%7B1%7D%7B2%7D+%5Ccos+2x+%2B+C++%7D+%5C%5C+%5C%5C+%5C%5C+%5Cmathsf%7By%27+%3D+2x+%2B+%5Cfrac%7B1%7D%7B2%7D+%5Ccdot+%28-%5Csin+2x+%5Ccdot+2%29%7D+%5C%5C+%5C%5C+%5C%5C+%5Cmathsf%7By%27+%3D+2x+-+%5Csin+2x%7D)
Verificando:
![\displaystyle \mathsf{\int 2x - \sin 2x \, \, dx} \\ \\ \\ \mathsf{2 \cdot \frac{x^2}{2}-\frac{1}{2} \cdot (-\cos 2x ) + C} \\ \\ \\ \mathsf{x^2+\frac{1}{2} \cos 2x + C} \displaystyle \mathsf{\int 2x - \sin 2x \, \, dx} \\ \\ \\ \mathsf{2 \cdot \frac{x^2}{2}-\frac{1}{2} \cdot (-\cos 2x ) + C} \\ \\ \\ \mathsf{x^2+\frac{1}{2} \cos 2x + C}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cmathsf%7B%5Cint+2x+-+%5Csin+2x+%5C%2C+%5C%2C+dx%7D+%5C%5C+%5C%5C+%5C%5C+%5Cmathsf%7B2+%5Ccdot+%5Cfrac%7Bx%5E2%7D%7B2%7D-%5Cfrac%7B1%7D%7B2%7D+%5Ccdot+%28-%5Ccos+2x+%29+%2B+C%7D+%5C%5C+%5C%5C+%5C%5C+%5Cmathsf%7Bx%5E2%2B%5Cfrac%7B1%7D%7B2%7D+%5Ccos+2x+%2B+C%7D)
Verificando:
Perguntas interessantes
Matemática,
11 meses atrás
Biologia,
11 meses atrás
Música,
1 ano atrás
Física,
1 ano atrás
Contabilidade,
1 ano atrás