Bisection Newton - Quadratures

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The problem states the following: Find with at least 10 digitis of precisión the roots of the following equation: $\int_x^{x^2} \!e^{-t^2}\,\mathrm{d}t = x^5 -3x^2 + 1 $ in the closed Interval [-1,1].

We know that we should use the Gaussian Method of Numerical Integration and the Bisection Method.

However we planned on using the Legendre-Gauss Quadrature, to first approximate the value of the integral, then use the bisection method to find the roots of the equation $\int_x^{x^2} \!e^{-t^2}\,\mathrm{d}t - x^5 + 3x^2 - 1 = 0 $

Is it an accurate approach? If so, when finding the weights and points to use in the Legendre-Gauss Quadrature we note that the Interval of the integral goes from $x$ to $x^2$. Does the LGQ method work still?

Thank you.

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It is still possible to use the quadrature method, call

$$ t = \frac{xu}{2}(x - 1) + \frac{x}{2}(x + 1) $$

so that $t(u = -1) = x$ and $t(u = +1) = x^2$, the integral then becomes

$$ \int_x^{x^2} e^{-t^2}{\rm d}t = \frac{x}{2}(x - 1)\int_{-1}^1 \exp\left[- \left(\frac{xu}{2}(x - 1) + \frac{x}{2}(x + 1)\right)^2 \right]{\rm d}u $$

That being said, it is actually not necessary to calculate the integral at all. Remember that

$$ {\rm erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}{\rm d}t $$

So that the problem reduces to

$$ \frac{\sqrt{\pi}}{2}[{\rm erf}(x^2) - {\rm erf}(x)] = x^5 - 3x^2 + 1 $$