How do you find #\int _ { 1} ^ { 16} \int _ { 1} ^ { 4} ( \frac { x } { y } + \frac { y } { x } ) d y d x#?
1 Answer
Explanation:
We want to evaluate:
# int_{1}^{16} int_{1}^{4} (x/y + y/x) \ dy \ dx #
We perform multiple integration by starting with the inner integral and integrate wrt to the variable of integration whilst treating other variables as constant. So:
# int_{1}^{16} int_{1}^{4} (x/y + y/x) \ dy \ dx #
# " "= int_1^16 { [xln|y|+y^2/(2x)]_(y=1)^(y=4) } #
# " "= int_1^16 { (xln4+16/(2x)) - (xln1+1/(2x)) } #
# " "= int_1^16 { xln4+16/(2x) - 1/(2x) } #
# " "= int_1^16 { xln4+15/(2x) } #
# " "= [ (x^2ln4)/2+15/2lnx ]_(x=1)^(x=16) #
# " "= ( (256ln4)/2+15/2ln16 ) - (ln4/2+15/2ln1) #
# " "= (256ln4)/2+15/2ln16 - ln4/2 #
# " "= (255*2ln2)/2+15/2*4ln2 #
# " "= 285ln2 #
