I created this question, and every other question in Multiple Integration, for my dissertation `Computer-Aided Assessment of Multiple Integration'.
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\n\\[ \\int_c^d \\int_a^b f(x,y) \\, \\mathrm{d}x \\, \\mathrm{d}y\\]
\n$a=\\,$[[0]] $b=\\,$[[1]]
\n$c=\\,$[[2]] $d=\\,$[[3]]
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\n\\[ f(x,y)=\\simplify{{s}x} + \\var{t} .\\]
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\nThis integral is type 2, because the order of integration is $y$, then $x$.
\nb)
\nSketch the graph, looks like this:
\nFrom looking at the graph we can see that the curve $y=\\simplify{{b}ln({c}x)+{d}}$, which was the upper limit for $y$ in type 2, is now the lower limit for $x$ in type 1. So we need to make $x$ the subject of the equation:
\n \\begin{align}
\\simplify{{b}ln({c}x)+{d}}=\\,& y \\\\
\\var{b}\\ln(\\var{c}x) =\\,& \\simplify{y-{d}} \\\\
\\ln(\\var{c}x) =\\,& \\simplify{(y-{d})/{b}} \\\\
\\var{c}x =\\,& e^{\\simplify{(y-{d})/{b}}} \\\\
x =\\,& \\frac{1}{\\var{c}}e^{\\simplify{(y-{d})/{b}}}.
\\end{align}
The upper limit is just the straight line: $x=\\var{q}$.
\n\nNow for $y$ the bottom limit is simply the line $y=\\var{a}$, and the top limit is the point of intersection between $y=\\simplify{{b}ln({c}x)+{d}}$ and $x=\\var{q}$, so $y=\\simplify{{b}ln{c*q}+{d}}$.
\n\nFinally, make sure you also swap $\\mathrm{d}x \\, \\mathrm{d}y$ since the integral is now in type 1.
\n\nAnswer:
\n\\[ \\int_\\var{a}^{\\simplify{{b}ln{c*q}+{d}}} \\int_{\\frac{1}{\\var{c}}e^{\\simplify[fractionNumbers]{(y-{d})/{b}}}}^{\\var{q}} f(x,y) \\, \\mathrm{d}x \\, \\mathrm{d}y \\]
\n\nc)
\n\\begin{align}
\\int_\\var{a}^{\\simplify{{b}ln{c*q}+{d}}} \\int_{\\frac{1}{\\var{c}}e^{\\simplify{(y-{d})/{b}}}}^{\\var{q}} \\simplify{{s}x}+\\var{t} \\, \\mathrm{d}x \\, \\mathrm{d}y
&=\\int_\\var{a}^{\\simplify{{b}ln{c*q}+{d}}} \\simplify{{s}/{2}x^2}+\\simplify{{t}x} \\, \\bigg|_{x=\\frac{1}{\\var{c}}e^{\\simplify{(y-{d})/{b}}}}^{x=\\var{q}} \\, \\mathrm{d}y
\\\\
&=\\int_\\var{a}^{\\simplify{{b}ln{c*q}+{d}}} \\simplify[fractionNumbers]{{s*q^2/2}}+\\var{t*q} - \\simplify[fractionNumbers]{{s/2}}\\cdot\\frac{1}{\\var{c}^2}e^{2\\simplify{(y-{d})/{b}}}-\\frac{\\var{t}}{\\var{c}}e^{\\simplify{(y-{d})/{b}}} \\, \\mathrm{d}y
\\\\
&=\\int_\\var{a}^{\\simplify{{b}ln{c*q}+{d}}} \\simplify[fractionNumbers]{{s*q^2/2+t*q}} - \\simplify[fractionNumbers]{{s/(2*c^2)}}e^{\\simplify{2*(y-{d})/{b}}}-\\simplify[fractionNumbers]{{t/c}}e^\\simplify{(y-{d})/{b}} \\, \\mathrm{d}y
\\\\
\\end{align}
\\begin{align}
\\int_\\var{a}^{\\simplify{{b}ln{c*q}+{d}}} \\simplify[fractionNumbers]{{s*q^2/2+t*q}} - \\simplify[fractionNumbers]{{s/(2*c^2)}}e^{\\simplify{2*(y-{d})/{b}}}-\\simplify[fractionNumbers]{{t/c}}e^\\simplify{(y-{d})/{b}} \\, \\mathrm{d}y
&=\\simplify[fractionNumbers]{{s*q^2/2+t*q}}y - \\simplify[fractionNumbers]{{b*s/(4c^2)}}e^{\\simplify{2*(y-{d})/{b}}}-\\simplify[fractionNumbers]{{b*t/c}}e^\\simplify{(y-{d})/{b}} \\bigg|_{y=\\var{a}}^{y=\\simplify{{b}ln{c*q}+{d}}}
\\\\
&=\\simplify[fractionNumbers]{{(s*q/2+t)*b*q}ln{c*q}} + \\simplify{({(s*q+2*t)*d*q/gcd0}/{2/gcd0})} - \\simplify[fractionNumbers]{{b*s/(4c^2)}}(\\var{c*q})^2-\\simplify[fractionNumbers]{{b*t/c}}\\cdot\\var{c*q} - \\simplify[fractionNumbers]{{(s*q/2+t)*a*q}} + \\simplify[fractionNumbers]{{b*s/(4c^2)}}\\simplify{e^({2*(a-d)}/{b})} + \\simplify[fractionNumbers]{{b*t/c}}\\simplify{e^({a-d}/{b})}
\\\\
&=\\simplify[fractionNumbers]{{(s*q/2+t)*b*q}ln{c*q}} + \\simplify{({k/gcd2}/{4/gcd2})} + \\simplify[fractionNumbers]{{b*s/(4c^2)}}\\simplify{e^({2*(a-d)}/{b})} + \\simplify[fractionNumbers]{{b*t/c}}\\simplify{e^({a-d}/{b})}
\\end{align}
Consider the integral
\n\\[ \\int_{\\frac{1}{\\var{c}}\\simplify{e^({a-d}/{b})}}^{\\var{q}} \\int_\\var{a}^{\\simplify{{b}ln({c}x)+{d}}} f(x,y) \\, \\mathrm{d}y \\, \\mathrm{d}x \\,.\\]
\nThe domain of integration is shown below.
\n{plot()}
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