integral (t^3=6t^2+4)dt from 0 to x
f''(x)=?
integral u^3(sqrtu+u^(1/5))du from 0 to 2
=?
what's with the + sign in the integral?
∫ u^3(√u + u^(1/5)) du
= e^(7/2) + u^(16/5) du
those are just normal power problems.
To find \(f''(x)\), we need to differentiate the given function \(f(x) = \int (t^3 - 6t^2 + 4) \, dt\) with respect to \(x\) twice.
Step 1: Evaluate the integral
First, let's find the antiderivative of the integrand \(t^3 - 6t^2 + 4\):
\(\int (t^3 - 6t^2 + 4) \, dt = \frac{1}{4} t^4 - 2 t^3 + 4t + C\), where \(C\) is the constant of integration.
Step 2: Calculate \(f(x)\)
We are given that \(f(x) = \int (t^3 - 6t^2 + 4) \, dt\). So, let's substitute \(x\) instead of \(t\) in the antiderivative:
\(f(x) = \frac{1}{4} x^4 - 2 x^3 + 4x + C\), where \(C\) is the constant of integration.
Step 3: Differentiate \(f(x)\)
Now, we differentiate \(f(x)\) with respect to \(x\) to find \(f'(x)\):
\(f'(x) = \frac{d}{dx} (\frac{1}{4} x^4 - 2 x^3 + 4x + C)\)
To differentiate, apply the power rule: Multiply each term by the corresponding exponent and subtract 1 from the exponent:
\(f'(x) = \frac{1}{4} \cdot 4x^{4-1} - 2 \cdot 3x^{3-1} + 4 \cdot 1x^{1-1}\)
Simplifying, we get:
\(f'(x) = x^3 - 6x^2 + 4\)
Step 4: Differentiate \(f'(x)\)
To find the second derivative \(f''(x)\), we differentiate \(f'(x)\) with respect to \(x\):
\(f''(x) = \frac{d}{dx} (x^3 - 6x^2 + 4)\)
Applying the power rule again, we have:
\(f''(x) = 3x^{3-1} - 6 \cdot 2x^{2-1}\)
Simplifying further:
\(f''(x) = 3x^2 - 12x\)
Therefore, \(f''(x) = 3x^2 - 12x\).