Answer:
The answer is "[tex]\frac{9}{2}[/tex]"
Step-by-step explanation:
If the given points are:
[tex]\to x^2 =8y\\\\\to -x+4y-4 =0\\\\[/tex]
please find the graph image in the attachment.
The intersection points are: [tex](-2,0.5) \ \ and \ \ (4,2)\\[/tex]
The formula for area:
[tex]Area= \int ( \text{upper curve - lower corve})\\\\[/tex]
[tex]= \int^4_{-2} [\frac{x+4}{4} - \frac{x^2}{8}] \ \ dx \\\\= \int^4_{-2} \frac{x+4}{4} \ dx - \int^4_{-2} \frac{x^2}{8} \ dx \\\\ = \frac{1}{4} \int^4_{-2} (x+4) \ dx- \frac{1}{8} \int^4_{-2} x^2 \ dx\\\\ = \frac{1}{4} (\frac {x^2}{2}+4x)^{4}_{-2}- \frac{1}{8} (\frac{x^3}{3})^{4}_{-2} \ dx\\\\ = \frac{1}{4} [(\frac {x^2}{2}+4x)^{4}_{-2}- \frac{1}{2} (\frac{x^3}{3})^{4}_{-2}] \ dx\\\\ = \frac{1}{4} [(\frac {x^2}{2}+4x- \frac{x^3}{6})^{4}_{-2}] \ dx\\\\= \frac{9}{2}\\[/tex]
