Question #281329

Evaluate the integral of 3 dx dy. Use inner limits=y^2 to 9, outer limits 0 to 3.



Determine the integral of xy dx dy. Using limits of x=0 to (y+1) and y=0 to 1.



Find the integral of (2-x-y) dx dy. Using limits of x=0 to (1-y)^0.5, and y=0 to 1.

1
Expert's answer
2021-12-20T17:08:01-0500

a)


03y293dxdy=303[x]9y2dy\displaystyle\int_{0}^3\displaystyle\int_{y^2}^{9}3dxdy=3\displaystyle\int_{0}^3[x]\begin{matrix} 9 \\ y^2 \end{matrix}dy

=303(9y2)dy=[27yy3]30=3\displaystyle\int_{0}^3(9-y^2)dy=[27y-y^3]\begin{matrix} 3 \\ 0 \end{matrix}

=81270=54=81-27-0=54

b)


010y+1xydxdy=01y[x22]y+10dy\displaystyle\int_{0}^1\displaystyle\int_{0}^{y+1}xydxdy=\displaystyle\int_{0}^{1}y[\dfrac{x^2}{2}]\begin{matrix} y+1 \\ 0 \end{matrix}dy

=1201(y3+2y2+y)dy=12[y44+2y33+y22]10=\dfrac{1}{2}\displaystyle\int_{0}^1(y^3+2y^2+y)dy=\dfrac{1}{2}[\dfrac{y^4}{4}+\dfrac{2y^3}{3}+\dfrac{y^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}

=12(14+23+12)=1724=\dfrac{1}{2}(\dfrac{1}{4}+\dfrac{2}{3}+\dfrac{1}{2})=\dfrac{17}{24}

c)


0101y(2xy)dxdy\displaystyle\int_{0}^1\displaystyle\int_{0}^{\sqrt{1-y}}(2-x-y)dxdy

=01[2xx22xy]1y0dy=\displaystyle\int_{0}^{1}[2x-\dfrac{x^2}{2}-xy]\begin{matrix} \sqrt{1-y} \\ 0 \end{matrix}dy

=01(21y1y2y1y)dy=\displaystyle\int_{0}^{1}(2\sqrt{1-y}-\dfrac{1-y}{2}-y\sqrt{1-y})dy

=[4(1y)3/23y2+y242(1y)5/25+2(1y)3/23]10=[-\dfrac{4(1-y)^{3/2}}{3}-\dfrac{y}{2}+\dfrac{y^2}{4}-\dfrac{2(1-y)^{5/2}}{5}+\dfrac{2(1-y)^{3/2}}{3}]\begin{matrix} 1 \\ 0 \end{matrix}

=12+14+23+25=4960=-\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{2}{3}+\dfrac{2}{5}=\dfrac{49}{60}


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