If f(x)=∫_0^x(t^3+6t^2+6)dt then f ' '(x)=

if f(x)=∫_0^x(t^3+6t^2+6)dt the f ' '(x)=? (That symbol is the integral sign 0 to x).

Also,
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. x+y^2=6 , x+y=0

Please show your work, thanks

f'(x) = x³ + 6x² + 6

f''(x) = 3x² + 12x

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Second question:

Graph:
http://www3.wolframalpha.com/input/?i=x%…

Integrate with respect to y

Find points of intersection:
x + y² = 6 -----> x = -y² + 6
x + y = 0 --------> x = -y

-y = -y² + 6
y² - y - 6 = 0
(y + 2) (y - 3) = 0
y = -2, y = 3

We integrate from y = -2 to y = 3
On this interval -y² + 6 > -y

A = ∫₋₂³ (-y² + 6 - (-y)) dy
A = ∫₋₂³ (-y² + y + 6) dy
A = (-y³/3 + y²/2 + 6y) |₋₂³
A = (-9 + 9/2 + 18) - (8/3 + 2 - 12)
A = 27/2 - (-22/3)
A = 81/6 + 44/6
A = 125/6 ≈ 20.8333

By the Fundamental Theorem of Calculus, the derivative of the integral is the thing under the integral sign. The derivative of the integral evaluated from 0 to x is the thing under the integral, evaluated at x.

So f'(x) = x^3 + 6x^2 + 6 and f''(x) is the derivative of that.