can't figure out how to solve this, completely forgot the math course Could you point by point how to do this?
\begin{gather}\int_0^x e^{-t^2} . (t^2-17t+72) \,dt\end{gather}
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To start this I would define $f(x) = \int^{x}_{0}{e^{-t^{2}}(t^2-17t+72)}$ and $g(t) = e^{-t^{2}}(t^2-17t+72)$ and recall that $max(f)$ occurs when $f’=0$. Then we can differentiate $f(x)$ $$f’(x) = \frac{d}{dx} \int^{x}_{0}{e^{-t^{2}}(t^2-17t+72)} = \frac{d}{dx} \int^{x}_{0}{g(t)} = \frac{d}{dx} [G(x)]^{x}_{0} = \frac{d}{dx} [G(x) - G(0)] = g(x)$$
Therefore $g(x) = 0$ which occurs when $e^{-x^{2}}=0$ (never) and when $x^2-17x+72=0$ so $x\in \{8,9\}$.
As the original equation is difficult to evaluate I would rely on the second derivative test. $$f’’(x)=g’(x)= -2xe^{-x^2} (x^2-17x+72)+e^{-x^2}(2x-17)$$ Note that $f’’(8) \lt 0 \lt f’’(9)$ Therefore $(8,f(8))$ is a maximum of the function.
EzTheBoss 2
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if we substitute it into the second order derivative, then at x = 8 we get -1/e^16 and at x=9 we get 1 / e ^ 18, doesn’t it turn out that the function is maximum at x = 9? – Денис Денис Mar 06 '23 at 19:46
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No, when the second derivative is negative we arrive at a maximum. This seems somewhat counterintuitive but it makes sense as at a maximum the slope is decreasing such that the curve becomes lower in value and the opposite is true of minimums. – EzTheBoss 2 Mar 06 '23 at 20:31
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