I wanna preface this: I never formally learned integrals and differentials
So, I'm currently helping a friend study for an exam, however... there's a specific exercise here, which is kinda out of my league:
With the function $h: \mathbb{R}\rightarrow\mathbb{R}$ defined as $h\left(x\right)\ =\ cx^{2}+2x+c^{2}$ calculate the smallest value resulting from the integral $\int_{0}^{2}h\left(x\right)dx$
I could use newtons method of just trying values, seeing when it reaches 0, but this seems extremely inefficient for this, and I assume there's better ways too...
If possible, maybe provide an explanation as to how the result would be calculated
Let's start by evaluating the Integral term by term: $$ \int_0^2 h(x)\text{d}x = \int_0^2 \left(cx^2+2x+c^2\right)\text{d}x = \left. c^2x+\frac{1}{3}cx^3+x^2 \right\vert_0^2 = \left( 2c^2 + \frac{1}{3}c\cdot2^3 + 2^2 \right) - \left( 0\cdot c^2 + \frac{1}{3}c\cdot0^3 + 0^2 \right) = 2c^2 + \frac{8}{3}c + 4 = f(c) $$ This results in a function that depends on $c$. Now we have to find the minimum of said function: $$ \min \left(\int_0^2 h(x)\text{d}x\right) = \min f(c) = \min\left(2c^2+\frac{8}{3}c+4\right) $$ This can be done using the derivative of $f$: $$ f'(c) = \frac{\text{d}}{\text{d}c} 2c^2+\frac{8}{3}c+4 = 4c + \frac{8}{3} $$ Solving $4c + \frac{8}{3} = 0$ yields $c_0 = -\frac{2}{3}$. To demonstrate that this is a minimum we can use the second derivative: $$ f''(c) = \frac{\text{d}}{\text{d}c} 4c + \frac{8}{3} = 4 \\ f''(c_0) = 4 \gt 0 \\ \Rightarrow \text{Minimum} $$ So the smallest value resulting from the given Integral is $f(c_0) = 2\left(-\frac{2}{3}\right)^2 + \frac{8}{3}\left(-\frac{2}{3}\right) + 4 = \frac{28}{9}$.