well, yeah.
f'' = 3x^2 + 2x
why did you take the antiderivative to go from f' to f''?
f(x)=∫(0,x) t^3+2t^2+2dt, and find f"(x).
my answer was:
f'(x)=x^3+x^2+2
f"(x)=x^4/4 + 2x^3/3 + 2x
it said its wrong.
f'' = 3x^2 + 2x
why did you take the antiderivative to go from f' to f''?
According to the Fundamental Theorem of Calculus, if F(x) is any antiderivative of f(x), then the definite integral of f(x) from a to b is equal to F(b) - F(a).
In this case, our function f(x) is defined as the definite integral of t^3 + 2t^2 + 2 with respect to dt, from 0 to x. So, let's first find an antiderivative F(x) of f(x).
To find F(x), we need to find the antiderivative of each term in the integrand. The antiderivative of t^3 is (1/4) t^4, the antiderivative of 2t^2 is (2/3) t^3, and the antiderivative of 2 is 2t. Therefore, an antiderivative of f(x) is:
F(x) = (1/4) x^4 + (2/3) x^3 + 2x + C,
where C is the constant of integration.
Now, to find f''(x), we need to take the derivative of F'(x). Taking the derivative of F(x) will give us:
F'(x) = d/dx [(1/4) x^4 + (2/3) x^3 + 2x + C]
= x^3 + 2x^2 + 2.
Therefore, f''(x) = F''(x) = d^2/dx^2 (x^3 + 2x^2 + 2).
Taking the second derivative of the function, we get:
f''(x) = d^2/dx^2 (x^3 + 2x^2 + 2) = 3x^2 + 4x.
So, your answer of f''(x) = x^4/4 + 2x^3/3 + 2x is incorrect. The correct second derivative is f''(x) = 3x^2 + 4x.