Question 574997
At x=0, y=1 for both functions, so they intersect at (0,1). The region would have vertical line x=1 and the two functions as boundaries. The region has the graph of {{{y=e^(7x)}}} as the upper boundary for y, and the graph of {{{y=e^(3x)}}} as its lower boundary throughout the [0,1] interval.
Integrating with respect to x would simply mean
{{{int( (e^(7x)-e^(3x)), dx, 0, 1 )}}}
If we try to integrate with respect to y, we need to do some calculations first:
{{{y=e^(3x)}}} and {{{x=1}}} intersect at (1,{{{e^3}}}), and
{{{y=e^(7x)}}} and {{{x=1}}} intersect at (1,{{{e^7}}})
The y values for the region range between 1 and {{{e^7}}}
The inverse of {{{y=e^(7x)}}} is {{{x=ln(y)/7}}} , which is the lower boundary for x values in the region.
On the other hand, the upper boundary for x values is {{{x=ln(y)/3}}} between 1 and {{{e^3}}}), and x=1 between {{{e^3}}}) and {{{e^7}}}).
So after all those calculations, we would end up with two integrals.
{{{int( (e^(7x)-e^(3x)), dx)=e^(7x)/7-e^(3x)/3}}}, so
{{{int( (e^(7x)-e^(3x)), dx, 0, 1 )=e^7/7-e^3/3-e^0/7+e^0/3=e^7/7-e^3/3-1/7+1/3=e^7/7-e^3/3+4/21}}} = approx. 150.157