Question

o valor da integral ∫³₀∫²₀ (4-y²)dy.dx é

Answer 1

Obs.: teorema fundamental do cálculo:

$\int^b_af(x)dx=F(x)\big|^b_a=F(b)-F(a).$

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$\int^3_0\int^2_0(4-y^2)dydx=$

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

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

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

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

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

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

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

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

$=\int^3_0(\frac{16}{3})\,dx=$

$=(\int\frac{16}{3}dx)\big|^3_0=$

$=\frac{16}{3}x\big|^3_0=$

$=\frac{16}{3}(3)-\frac{16}{3}(0)=$

$=16-\frac{0}{3}=$

$=16.$

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