:browse confirm e Area Solution 12

SOLUTION 12: $\ \$ If $x = \displaystyle{ \ln y - { y^2 \over 8 } }$ for $1 \le y \le 2$, then $$\displaystyle{ { dx \over dy} = { 1 \over y } - { 2y \over 8 } }$$ $$= \displaystyle{ { 1 \over y } - { y \over 4 } }$$ so that $$ARC = \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big( { 1 \over y } - { y \over 4 } \Big)^2 } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big( { 1 \over y^2} - { 1 \over 2 } + { y^2 \over 16 } \Big) } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ { 1 \over y^2} + { 1 \over 2 } + { y^2 \over 16 } } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ { 1 \over y^2 }{16 \over 16} + { 1 \over 2 }{ 8y^2 \over 8y^2 } + { y^2 \over y^2 }{ y^2 \over 16 } } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ { { y^{4} + 8y^2 + 16 } \over 16y^2 } } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \sqrt{ y^{4} + 8y^2 + 16 } \over \sqrt{ 16y^2 } } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \sqrt{ (y^{2} + 4)^2 } \over \sqrt{ (4y)^2} } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { { y^{2} + 4 } \over 4y } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \Big( { y^{2} \over 4y } + { 4 \over 4y} \Big) } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \Big( (1/4)y + { 1 \over y } \Big) } \ dy }$$ $$= \displaystyle{ { \Big( (1/4){y^{2} \over 2} + \ln |y| } \Big) \ \Big\vert_{1}^{2} }$$ $$= \displaystyle{ \Big( { {y^{2} \over 8} + \ln |y| } \Big) \ \Big\vert_{1}^{2} }$$ $$= \displaystyle{ \Big( { {(2)^{2} \over 8} + \ln |2| } \Big) - \Big( { {(1)^{2} \over 8} + \ln |1| } \Big) }$$ $$= \displaystyle{ {4 \over 8} + \ln 2 - { 1 \over 8 } - 0 }$$ $$= \displaystyle{ {3 \over 8} + \ln 2 }$$