1) f(x) = sinx m [0; 2л};
2) f(x) = cosx m [-ё:%"‚.
[m]S=∫ ^{π}_{0}(sinx-0)dx+ ∫ ^{2π}_{π}(0-sinx)dx=[/m]
[m]=(-cosx)| ^{π}_{0}+(cosx)|^{2π}_{π}=(-cosπ-(-cos0))+(cos2π-cosπ)=(-1(-1)+1)+(1-(-1)=2+2=4[/m]
или
[m]S=2\cdot ∫ ^{π}_{0}sinxdx=2\cdot (-cosx)| ^{π}_{0}=2\cdot (-cosπ+cos0)=2\cdot (-(-1)+1)=4[/m]
2)
[m]S=∫ ^{\frac{π}{2}}_{-\frac{π}{2}}(cosx-0)dx+ ∫ ^{\frac{3π}{2}}_{\frac{π}{2}}(0-cosx)dx=[/m]
[m]=(sinx)| ^{\frac{π}{2}}_{-\frac{π}{2}}-(sinx)|^{\frac{3π}{2}}_{\frac{π}{2}}=sin\frac{π}{2}-sin(-\frac{π}{2})-(sin\frac{3π}{2}-sin\frac{π}{2})=(1-(-1))-(-1-1)=2+2=4[/m]
или
[m]S=2\cdot ∫ ^{\frac{π}{2}}_{-\frac{π}{2}}cosxdx=2\cdot (sinx)| ^ {\frac{π}{2}}_{-\frac{π}{2}}=[/m]
[m]=2\cdot (-sin\frac{π}{2}-sin(-\frac{π}{2}))=2\cdot (1-(-1))=4[/m]