y=-2x+8 & y= 12 over the interval -1<x<2

[stephanie953]

y=-2x+8 and y= 12 over the interval -1<x<2. Using { for the integral sign

{(-2x + 8) dx
-2{x dx + 8{ dx
-2 x^2/2 + 8x
-x^2 + 8x
(-2^2 + 8(2)) - (-1^2+8(-1))
20 - (-7) = 27
27 - 12 = 15

Is this done correctly. Or is there a different way to handle the 2 different y's? Thank you.

 

[fasteddie65]

"y=-2x+8 and y= 12 over the interval -1<x<2. Using { for the integral sign"

First of all, you need to find the intersetion of the two curves; in this case, lines.

-2x + 8 = 12
-2x = 4
x = -2

We have two lines intersecting at (-2, 12)

The lines x = -1, x = 2, y = 12 and y = -2x + 8 form a trapezoid.

The integral finds the area under a curve, so find the integral of 12 - (-2x + 8) = 4 + 2x, from x = -1 to x = 2.

 

[stephanie953]

Thanks so much. That was very helpful. I got the same answer with this change in work.

4{dx + 2{x dx
4x| + x^2|
(4(2) + 2^2) - (4(-1) + -1^2)
12 + 3 = 15