Answer: hello your question is incomplete attached below is the complete question
answer :
a) [tex]\int\limits^3_0 {(10-x)-(x+4)} \, dx[/tex] ( option D )
b) A = 1/2 (6)(3) ( option B )
c) Area of shaded region = 9
Step-by-step explanation:
a) Using integration with respect to x
Area = [tex]\int\limits^7_4 {(y-4)} \, dy + \int\limits^a_7 {(10-y)} \, dy[/tex] ( note a = 10 )
= y^2/2 - 4y |⁷₄ + 10y - y^2/2 |¹⁰₇
= 33/2 - 12 + 30 - 51/2 = 9
hence the best integral from the options attached is option D
[tex]\int\limits^3_0 {(10-x)-(x+4)} \, dx[/tex]
= [ 10x - x^2 /2 - x^2/2 - 9x ] ³₀
= 30 - 9/2 - 9/2 - 12 = 9
b) Using Geometry
Area = 1/2 * base * height
= 1/2 * 6 * 3
= 9
