Find area underneath the curve question HELP!?

I assume the area in question is bounded above by y = n(16 - x^2) and below by the x-axis???

1. Before you create an integral expression, you have to know the interval over which you're integrating. To find your limits of integration, find the roots of the quadratic. [Yes, it's a quadratic...

y = n(16 - x^2) = (-n)x^2 + 0x + 16n.] Use the Quadratic formula.

y = n(16 - x^2)

y = -n(x^2 - 16)

0 = -n(x^2 - 16)

0 = x^2 - 16

0 = (x-4)(x+4)

So you're integrating over the interval [-4,4], and your integral expression becomes:

integral[-4,4] { n(16 - x^2) dx } =

n * integral[-4,4] { (16 - x^2)dx }

2. I assume the "shaded area" is the area bounded by the x-axis and y = n(16 - x^2)???

integral[-4,4] { n(16 - x^2)dx } =

n * integral[-4,4] { (16 - x^2)dx } =

n (16x - (1/3)x^3), evaluated on [-4,4] =

n[((16)(4) - (1/3)(4^3)) - ((16)(-4) - (1/3)(-4)^3)] =

n[(64 - 64/3) - (-64 + 64/3)] =

n(128/3 + 128/3) =

n(256/3)

76 = n(256/3)

228/256 = n

57/64 = n

y=ok cos 3x considering curve cuts x axis then y = 0 ok cos(3x) = 0 ok isn't 0, consequently cos(3x) =0 or 3x = (2n+a million)(pi/2) x = (2n+a million)(pi/6) ...............Ans Therefor factor of A(pi/6,0) ...............ans whilst the factor is on y axis positioned x = 0 y = ok using fact cos (0) = a million and cos (6pi) = -a million then y = -ok then factor on y-axis = B(0,ok)