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Evaluate the integral.

010x20x+y(2xyz)dz dy dx\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { x ^ { 2 } } \int _ { 0 } ^ { x + y } ( 2 x - y - z ) d z\ d y\ d x

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Answered 1 year ago
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010x20x+y(2xyz)dzdydx=010x2(0x+y(2xyz)dz)dydx=010x2(0x+y2xdz0x+yydz0x+yzdz)dydx=010x2(2xz0x+yyz0x+yz220x+y)dydx=010x2(2x(x+y0)y(x+y0)((x+y)22022))dydz=010x2(2x2+2xyxyy2x22xyy22)dydx=010x23x23y22dydx=12010x2(3x23y2)dydx=1201(0x23x2dy0x23y2dy)dx=1201(3x2y0x23y330x2)dx=1201(3x2(x20)3((x2)33033))dx=1201(3x43x63)dx=1201(3x4x6)dx=12(013x401x6)=12(3x5501x7701)=12(3(155055)177+077)=12(3517)=835\begin{align*} \int_{0}^{1} \int_{0}^{x^{2}} \int_{0}^{x+y} (2x - y - z) dz dy dx &= \int_{0}^{1} \int_{0}^{x^{2}} \left( \int_{0}^{x+y} (2x - y - z) dz\right) dy dx\\ &= \int_{0}^{1} \int_{0}^{x^{2}} \left( \int_{0}^{x+y} 2x dz - \int_{0}^{x+y} y dz - \int_{0}^{x+y} z dz \right) dy dx \tag{Divide the integral in three integrals.}\\ &= \int_{0}^{1} \int_{0}^{x^{2}} \left( 2x\left.z \right|_{0}^{x+y} - y\left.z \right|_{0}^{x+y} - \left.\frac{z^{2}}{2} \right|_{0}^{x+y}\right) dydx \tag{Integrate in relation to $z$.}\\ &= \int_{0}^{1} \int_{0}^{x^{2}} \left( 2x(x+y-0) - y(x+y - 0) - \left(\frac{(x+y)^{2}}{2} - \frac{0^{2}}{2}\right)\right) dy dz \tag{Compute the boundaries.}\\ &= \int_{0}^{1} \int_{0}^{x^{2}} \left( 2x^{2} + 2xy - xy - y^{2} - \frac{x^2}{2}-xy-\frac{y^2}{2} \right) dy dx \tag{Simplify.}\\ &= \int_{0}^{1} \int_{0}^{x^{2}} \frac{3x^2-3y^2}{2} dy dx \\ &= \frac{1}{2} \int_{0}^{1} \int_{0}^{x^{2}} (3x^2-3y^2) dy dx \tag{Take the constant $\frac{1}{2}$ out.}\\ &= \frac{1}{2} \int_{0}^{1} \left( \int_{0}^{x^{2}} 3x^2 dy - \int_{0}^{x^{2}} 3y^2 dy\right) dx \tag{Divide the integral in two integrals.}\\ &= \frac{1}{2} \int_{0}^{1} \left( 3x^{2} \left.y \right|_{0}^{x^{2}} - 3\left. \frac{y^{3}}{3} \right|_{0}^{x^{2}} \right) dx \tag{Integrate in relation to $y$.}\\ &=\frac{1}{2} \int_{0}^{1} \left( 3x^{2} (x^{2} - 0) - 3 \left(\frac{(x^{2})^{3}}{3} - \frac{0^{3}}{3} \right) \right) dx \tag{Compute the boundaries.}\\ &= \frac{1}{2} \int_{0}^{1} \left( 3x^{4} - \frac{3x^{6}}{3} \right) dx \tag{Simplify.}\\ &= \frac{1}{2} \int_{0}^{1} (3x^{4} - x^{6}) dx\\ &= \frac{1}{2} \left( \int_{0}^{1} 3x^{4} - \int_{0}^{1} x^{6}\right) \tag{Divide the integral in two integrals.}\\ &= \frac{1}{2} \left( 3 \left. \frac{x^{5}}{5}\right|_{0}^{1} - \left. \frac{x^{7}}{7}\right|_{0}^{1} \right) \tag{Inetegrate in relation to $x$.}\\ &= \frac{1}{2} \left( 3 \left(\frac{1^{5}}{5} - \frac{0^{5}}{5} \right) - \frac{1^{7}}{7} + \frac{0^{7}}{7} \right) \tag{Compute the boundaries.}\\ &= \frac{1}{2} \left( \frac{3}{5} - \frac{1}{7}\right) \tag{Simplify.}\\ &= \textcolor{#c34632}{\frac{8}{35}} \end{align*}

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