7.1 prob 4 Consider the region between the graphs x=y^2-6 and x+5y=8

Question

where a= -7 and b=2 and h(y)=??

The area is 121.5

**I got all of the answer except h(y).. not sure why that part of my answer is wrong when the rest is correct. I got h(y)=(y^2-5)-(9-5y)

Answer 1

Looks like you miscopied the problem as shown below in bold and italic...

integral ( y^2 -6 - (-5y+8) ) =

integral ( y^2 - 6 + 5y - 8) =

integral ( y^2 + 5y - 14) =

y^3/3 + 5y^2/2 - 14y

since y ranges from a=-7 to b=2 , we must switch these limits of integration.

But then we must take the OPPOSITE of the integral. That is,

integral from a to b = - integral from b to a

So the limits are 15 and 1/3 = 15.3 and 106 and 1/6 = 106.16

The total area behind the curve is then 121.5 = 121 and 1/2