How do you do this Integration problem, please? I keep getting 1440 but the answer is wrong.?

Question

Q: Find the integral by interpreting it in terms of its areas.

The integral from -4 to 4 of (16-(x^2))^(1/2), or the square root of (16-(x^2)).

Thanks!

Answer 1

Since the function is only evaluated from -4 to 4 and has a range from 0 to 4 in that domain, 1440 seems way out of line. This looks like a listed integral. Check a calc book or wikipedia.

Answer 2

To interpret this integral in terms of the area square both sides of y = sqrt(16-x^2) to get y^2 + x^2 = 16 (after adding x^2 to both sides). This is the the equation of a circle with center at origin and radius 4. But we want only positive values for y so we are looking at the half circle with radius 4. The integra then is just the area of this half circle which would be half of pir^2 or A = pi (4^2)/2 = 8pi.

Answer 3

integral of {16 - (x^2)} ^ (1/2) = [{16 - (x^2)} ^ (3/2)] / {(3/2) * -2x} = - 1/3 * [{16 - (x^2)} ^ (3/2)]

the value from 4 to -4 = - 1/3 * [{16 - (4^2)} ^ (3/2)] - - 1/3 * [{16 - ((-4)^2)} ^ (3/2)] = 0

Answer 4

x = 4sin u
dx = 4cos u
u = arcsin (x/4)

Integral (-pi/2 to pi/2) sqrt (16 - (4sin u)^2) cos u du =
Integral (-pi/2 to pi/2) 4cos u * cos u du =
4(Integral (-pi/2 to pi/2) cos u * cos u du)

cos u = v
cos u du = dw
-sin u du = dv
sin u = w

4[(sin 2u) / 2 - Integral -sin u * sin u du] =
4[(sin 2u) / 2 + Integral sin^2 u du] =
4[(sin 2u) / 2 + u - Integral cos^2 u du] =
4(Integral cos^2 u du)

4(Integral cos^2 u du) = (sin 2u) + 2u + C

Integrate from -pi/2 to pi/2.
Calculate by yourself.