the fundamental theorem of calculus,
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.
well, yeah.
f'' = 3x^2 + 2x
why did you take the antiderivative to go from f' to f''?
To find the second derivative of the function f(x) = ∫(0,x) t^3 + 2t^2 + 2 dt, we need to apply the Fundamental Theorem of Calculus.
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.