calcccc help! i dont get itt?

Question

Find the area between a parabola y=-4 x^2 + 0 x +64 and the x-axis.
HINT: Find the intersection points of the parabola with the x-axis first.


Area=

Answer 1

to find where the parabola intersects the x-axis all you have to do is set y to zero.

0 = -4x^2 + 0x + 64 <-- 0 times x = 0 so the equation is really
0 = -4x^2 +64 <-- bring the 64 to the other side
-64 = -4x^2 <-- divide both sides by -4
16 = x^2 <-- take the square root of both sides to isolate x
4 = x
so the parabola crosses the x axis at x = 4

no I'm guessing all you have to do is find the integral of
4x^2 + 64 which is pretty easy.

integral (4x^2 +64)dx is the same as:
integral 4x^2dx + integral 64dx

I'm guessing you know how to do simple integrals like this, but if you don't all you have to remember for ones like these to is that if you take the integral of a number then the derivative of that number, you'll get that number again.

eg.
integral of 4x^2 = 4x^3/3
and the derivative of 4x^3/3 = 4x^2

the integral of 64 = 64x
and the derivative of 64x is 64

anywho, I think the area is the addition of the two integrals with x equalling 4 (but I'm can't remember for sure so don't take my word completely).

eg. 4(4)^3/3 +64(4) = Area

PS. The guy right above me is right..it's actually the integral of 4x^2 + 64 from -4 to +4

Answer 2

The hint says find the intersection points first. Those are found by setting y equal to zero and solving the resulting quadratic:

0 =-4 x^2 + 0 x +64

0 = x^2 - 16

x = +/- 4

So you now know you must integrate the given function from x = -4 to x = +4. The antiderivative of your function is - ( 4 / 3 ) x^3 + 64 x + C.

Answer 3

0 = y = -4x² + 64

4x² = 64

x² = 64/4

x = ±√(64/4)


Integrate (-4x² + 64)dx on the interval -4 to +4.

Evaluate (-4xxx)/3 + 64x with -4 and +4, and find the difference.