Obs.: teorema fundamental do cálculo:

[tex]\int^b_af(x)dx=F(x)\big|^b_a=F(b)-F(a).[/tex]

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[tex]\int^3_0\int^2_0(4-y^2)dydx=[/tex]

[tex]=\int^3_0\Big[\int^2_0(4-y^2)dy\Big]dx=[/tex]

[tex]=\int^3_0\Big[\int(4-y^2)dy\big|^2_0\,\Big]dx=[/tex]

[tex]=\int^3_0\Big[(\int4dy-\int y^2dy)\big|^2_0\,\Big]dx=[/tex]

[tex]=\int^3_0\Big[(4y-\frac{y^3}{3})\big|^2_0\,\Big]dx=[/tex]

[tex]=\int^3_0\Big[(4(2)-\frac{2^3}{3})-(4(0)-\frac{0^3}{3})\big|^2_0\,\Big]dx=[/tex]

[tex]=\int^3_0\Big[(8-\frac{8}{3})-(0-\frac{0}{3})\,\Big]dx=[/tex]

[tex]=\int^3_0\Big[(\frac{24-8}{3})-(0-0)\,\Big]dx=[/tex]

[tex]=\int^3_0\Big[\frac{16}{3}-0\,\Big]dx=[/tex]

[tex]=\int^3_0(\frac{16}{3})\,dx=[/tex]

[tex]=(\int\frac{16}{3}dx)\big|^3_0=[/tex]

[tex]=\frac{16}{3}x\big|^3_0=[/tex]

[tex]=\frac{16}{3}(3)-\frac{16}{3}(0)=[/tex]

[tex]=16-\frac{0}{3}=[/tex]

[tex]=16.[/tex]

[tex]\underline{\boxed{\boxed{\int^3_0\int^2_0(4-y^2)dydx=16}}}\,.[/tex]