Answer to Question #150487 in Calculus for sai

Question

sketch the region of integration and write an equivalent double integral with order of integration reversed and then calculate it.

integral 0 to 3 integral 1 to e^y (x+y) dx dy

Answer 1

The region of integration:

Double integral with order of integration reversed:

$\int_1^{exp[3]} \int_{lnx}^{3} (x+y)dy dx=\int_1^{exp[3]} \int_{lnx}^{3} (x+y)dy dx$ =

$\int_1^{exp[3]} (0.5*(9-ln^2(x))-x(lnx-3)) dx$=

$\int_1^{exp[3]} (-0.5(lnx-3)(2x+lnx+3)) dx$ =

$-0.5\int_1^{exp[3]} (-6x+ln^2x+2x*lnx-9) dx$ =

$-0.5\int_1^{exp[3]} (ln^2x) dx-\int_1^{exp[3]} (x*lnx) dx+3\int_1^{exp[3]} (x) dx+9/2*\int_1^{exp[3]} (1) dx$

$(-0.5*x*ln^2x)\vert_1^{e^3}+\int_1^{exp[3]} (lnx) dx-\int_1^{exp[3]} (x*lnx) dx+3\int_1^{exp[3]} (x) dx+9/2*\int_1^{exp[3]} (1) dx$

$-9/2*e^3+(x*lnx)\vert_1^{e^3}+7/2\int_1^{exp[3]} (1) dx-\int_1^{exp[3]} (x*lnx) dx+3\int_1^{exp[3]} (x) dx$

$-3/2*e^3+7/2\int_1^{exp[3]} (1) dx-\int_1^{exp[3]} (x*lnx) dx+3\int_1^{exp[3]} (x) dx$

$-3/2*e^3+7x/2\vert_1^{e^3}-\int_1^{exp[3]} (x*lnx) dx+3\int_1^{exp[3]} (x) dx$

$-3/2*e^3+7/2(e^3-1)-\int_1^{exp[3]} (x*lnx) dx+3\int_1^{exp[3]} (x) dx$

$-3/2*e^3+7/2(e^3-1)+(-0.5x^2*lnx)\vert_1^{e^3}+7/2\int_1^{exp[3]} (x) dx$

$-3/2*e^3-3*e^6/2+7/2(e^3-1)+7/2\int_1^{exp[3]} (x) dx$

$-3/2*e^3-3*e^6/2+7/2(e^3-1)+7x^2/4\vert_1^{e^3}$

$-3/2*e^3-3*e^6/2+7/2(e^3-1)+7/4*(e^6-1)$