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

  • 6
    $\begingroup$ Have you tried evaluating the integral? $\endgroup$
    – Boxonix
    Commented Mar 19, 2023 at 17:58
  • 2
    $\begingroup$ You are expected to evaluate the integral leaving $c$ as a parameter. You will get an expression involving $c$. Now find the minimum of that expression, say by differentiating with respect to $c$ and setting that to $0$. $\endgroup$ Commented Mar 19, 2023 at 18:18
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    $\begingroup$ For the record, Newton's method is not "just trying values". $\endgroup$
    – lulu
    Commented Mar 19, 2023 at 18:22
  • $\begingroup$ @lulu I am aware, newtons method is choosing a starting position and then reevaluating the function with a new input value going up/down depending on the last result value until a sufficiently close enough value is found. "Just trying values" wasn't for describing newtons method, but rather trying pseudo-random input values - this would have no guarantee to result in the desired output, hence why I said there's better ways $\endgroup$
    – Lilly
    Commented Mar 19, 2023 at 18:36
  • 1
    $\begingroup$ But this is not a problem set up for Newton's method (which is a root finding algorithm). Just compute the integral, and the answer drops out quickly. $\endgroup$
    – lulu
    Commented Mar 19, 2023 at 18:38

1 Answer 1


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 $f(c_0) = 2\left(-\frac{2}{3}\right)^2 + \frac{8}{3}\left(-\frac{2}{3}\right) + 4 = \frac{28}{9}$.

  • 1
    $\begingroup$ Second derivative test tests for local extrema, not global. It's clear that this is indeed global minimum since we have quadratic function with positive leading coefficient, but the way you wrote it might be misleading. $\endgroup$
    – Ennar
    Commented Mar 19, 2023 at 18:38
  • $\begingroup$ @Ennar You are completely right, I rewrote that part. $\endgroup$
    – Nerrit
    Commented Mar 19, 2023 at 18:40
  • $\begingroup$ Thanks so much for this, as I said, I never formally learned integrals and derivates, so there's a lot of knowledge I'm missing on that topic. You also explained it really well! $\endgroup$
    – Lilly
    Commented Mar 19, 2023 at 18:53

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