Verified answer

top minus bottom:

Sketching:

http://www.wolframalpha.com/input/?i=Plot%5B%7B-4+...

two parts:

limits:

y = y

-4x = 4

x = -1

y = y

√(x) = 4

x = 16

0. . . .. .. .. .. .16

∫ (4 - -4x) dx + ∫ ( 4 - √(x) ) dx

-1. .. .. .. .. .. .0

0. . . .. .. .. .. .16

∫ (4 + 4x) dx + ∫ ( 4 - √(x) ) dx

-1. .. .. .. .. .. .0

. . . . . . . . . .0. . . .. .. . . . . .. . . . . . . . . . 16

(4x + 2x^2 ) ] + ( 4x - x^(1/2 + 1)/(1/2 + 1) ) ]

. . . . . . . . . -1. .. .. .. .. . . . . . . . . . . . . . . 0

. . . . . . . . . .0. . . .. .. . . . . .. . .. . 16

(4x + 2x^2 ) ] + ( 4x - x^(3/2)/(3/2) ) ]

. . . . . . . . . -1. .. .. .. .. . . . . . . . . 0

. . . . . . . . . .0. . . .. .. . . . . .. . .. . 16

(4x + 2x^2 ) ] + ( 4x - (2/3) * x^(3/2) ) ]

. . . . . . . . . -1. .. .. .. .. . . . . . . . . 0

(4 * (0 - -1) + 2 * ( 0^2 - (-1)^2 ) ) + ( 4 * (16 - 0) - (2/3) * ( 16^(3/2) - 0^(3/2) )

(4 * (0 + 1) + 2 * ( 0 - 1 ) ) + ( 4 * 16 - (2/3) * ( 64 - 0 )

(4 * 1 - 2 ) + ( 64 - (128/3) )

(4 - 2 ) + (64/3)

2 + (64/3)

70/3

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I got 70/3, not sure if it's right.

Once you plot out the three lines, the area bounded by the lines is neatly separated into two sections, separated by the y axis.

The left part is a triangle, and we can find the intersection point, the leg lengths of a right triangle, and hence we can find the triangle bounded by y = 4, y = -4x, and the y axis.

The right part is a little harder. If you draw a line perpendicular to where the y = 4 line and y=sqrt(x) line vertically down to touch the x axis, it should touch at 16 because in y = sqrt(x), when y is 4 x is 16.

From there, we have a rectangle. I'm not sure how to find the area of the TOP part that we need directly, but I can find the bottom part and subtract it from the total area of the rectangle.

Using integrals, we have the area from 0 to 16 with the formula x^1/2. That gives us (x^3/2)/(3/2) or (2x^3/2)/3. Plugging in 0 and 16, subtracting, and taking the absolute value we get 128/3. If the entire area of the rectangle is 64, or 192/3, then if you subtract that part, we get the area of the right half to be 64/3.

Add both together, and you get 70/3. I believe that this is right, although I apologize if this is wrong, as I have only had Algebra and very basic calculus.