Find the total area between y=x+14 and y=4x+3 between x = 0 and x =2.

Question

Find the total area between y=x+14 and y=4x+3 between x = 0 and x =2.

Answer 1

overkill! WE DON'T NEED ANY CALCULUS AT ALL FOR THIS ONE.

 

Consider trapezoid #! with vertical base of length 14, on the y-axis from the origin to the y-intercept of y = x + 14.

Its other base runs from (2,0) to (2,16) so the area of this trapezoid is "average base" times height, or 15 times 2, i.e. area is 30.

 

Subtract from that, the area of trapezoid # 2, which has smaller base running from the origin to (0,3) and larger base running from (2,0)  to (2,11), so its avetrage base is 7 and its height is also 2 so this area is 14.

 

The difference 30 - 14 is the answer, area = 16.

 

And if you want to use calculus, a double integral is not necessary, either.

Answer 2

OK. Double integral time.

 

∫∫ dy dx (as y goes from x+14 to 4x+3)(as x goes from 0 to 2)

 

= ∫ [-3x + 11] dx (as x goes from 0 to 2)

 

= (-3/2)x² + 11x (as x goes from 0 to 2)

 

= 22 - 6 = 16

 

Sanity check: I ran this through Wolfram Alpha and got the same answer.

 

Cheers!