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(a) We proceed to evaluate the double-integral:

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\\[\\begin{eqnarray*} I&=&\\int^\\var{a}_1 \\int^\\var{b}_0 \\left(\\var{c}+\\simplify[std]{{4*d}xy} \\right)  dx  dy \\\\ &=& \\int^\\var{a}_1 \\left[\\simplify[std]{{c}x+{2*d}*y*x^2} \\right]^\\var{b}_0  dy \\\\ &=&\\int^\\var{a}_1 \\left(\\simplify[std]{{c*b}+{2*d*b^2}*y} \\right)  dy \\\\ &=& \\left[\\simplify[std]{{c*b}y+{d*b^2}*y^2} \\right]^\\var{a}_1  dy \\\\ &=&\\simplify[std]{{c*b*a}+{d*b^2*a^2}-{c*b}-{d*b^2}} \\\\ &=&\\simplify[std]{{(c*b*a)+(d*b^2*a^2)-(c*b)-(d*b^2)}}\\end{eqnarray*}\\]

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(b) \\[\\begin{eqnarray*} I&=&\\int^\\pi_0 \\int^\\var{h}_0 \\simplify[std]{y^{f}sin({g}x)} dy  dx \\\\ &=& \\int^\\pi_0 \\left[\\simplify[std]{(1/{f+1})*y^{f+1}*sin({g}x)}\\right]^\\var{h}_0 dx \\\\ &=& \\int^\\pi_0 \\simplify[std]{({h}^{f+1}/{f+1})*sin({g}x)}  dx  \\\\ &=& \\simplify[std]{({h}^{f+1}/{f+1})}\\left[\\simplify[std]{-1/{g}*cos({g}x)}\\right]^\\pi_0  \\\\ &=& -\\simplify[std]{({h}^{f+1}/{g*(f+1)})} \\left(\\simplify[std]{{(-1)^g}}-1 \\right) \\\\ &=& \\simplify[fractionnumbers]{{-{h}^({f+1})*((-1)^{g}-1)/({g*(f+1)})}}\\end{eqnarray*}\\]

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\\[I=\\int^\\var{a}_{y=1} \\int^\\var{b}_{x=0} \\left(\\var{c}+\\simplify[std]{{4*d}xy} \\right)  dx\\,  dy \\]

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$I=\\;\\;$[[0]]

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Input all numbers in your answer as integers or fractions, not as decimals.

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\\[I=\\int^\\pi_{x=0} \\int^\\var{h}_{y=0} \\simplify[std]{y^{f}sin({g}x)} dy \\, dx \\]

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$I=\\;\\;$[[0]]

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Evaluate the following double integrals.

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Input your answer as an integer or a fraction, not as a decimal.

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Double integrals (2) with numerical limits

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