Use the interval addition Rule?

Question

Use the interval addition Rule to evaluate the definite integral to find the area between the graph of g and the x-axis,

evaluate from -4 to 1
g(x) dx,where g(x) = 8 - x^2, x < -2 ;
x + 8, x is greater than or equal to -2

Please show your work

(a) 125 / 2

(b) 119 / 6

(c) 31

(d) 61 / 2

Answer 1

The g(x)=8-x^2 will intercept x axis at x=-sqrt(8) and x= +sqrt(8)
the g(x)=8+x will intercept x axis at x=-8

The area bounded by the functions and x axis is a sum two integrals
1. |(8-x^2)dx from -sqrt(8) to -2
2. |(8+x)dx from -2 to 1

The answer I got was ((8+64(2)^.5))/6

Answer 2

You need to evaluate the integral of 8 - x^2 from -4 to -2 and the integral of x + 8 from -2 to 1

int(8 - x^2) = 8x - (x^3)/3
To evaluate from -4 to -2, plug in those values for x and subtract:
(8(-2) - ((-2)^3)/3 - [(8(-4) - ((-4)^3)/3]
-16 - (-8/3) - [-32 - (-64)/3]
-48/3 + 8/3 + 96/3 - 64/3
-8/3

int(x +8) = (x^2)/2 + 8x
Evaluated from -2 to 1, this gives:
(1^2)/2 + 8(1) - [(-2)^2/2 + 8(-2)]
1/2 + 8 - 4/2 + 16
1/2 + 16/2 - 4/2 + 32/2
45/2

Now add the two results:
-8/3 + 45/2 = -16/6 + 135/6 = 119/6

Answer choice B is correct.