Question

Calculate by double integration the area of the bounded region determined by the given pairs of curves. g

Answer 1

The answer is "
`(9)/(2)`"

Explanation:

If the given points are:

`\to x^2 =8y\\\\\to -x+4y-4 =0\\\\`

please find the graph image in the attachment.

The intersection points are:
`(-2,0.5) \ \ and \ \ (4,2)\\`

The formula for area:

`Area= \int ( \text{upper curve - lower corve})\\\\`

`= \int^4_(-2) [(x+4)/(4) - (x^2)/(8)] \ \ dx \\\\= \int^4_(-2) (x+4)/(4) \ dx - \int^4_(-2) (x^2)/(8) \ dx \\\\ = (1)/(4) \int^4_(-2) (x+4) \ dx- (1)/(8) \int^4_(-2) x^2 \ dx\\\\ = (1)/(4) (\frac {x^2}{2}+4x)^(4)_(-2)- (1)/(8) ((x^3)/(3))^(4)_(-2) \ dx\\\\ = (1)/(4) [(\frac {x^2}{2}+4x)^(4)_(-2)- (1)/(2) ((x^3)/(3))^(4)_(-2)] \ dx\\\\ = (1)/(4) [(\frac {x^2}{2}+4x- (x^3)/(6))^(4)_(-2)] \ dx\\\\= (9)/(2)\\`