Solve this Integral Without a Calculator

Solution 1:

Too long for a comment.

Let me suppose that there is not typo and that we want to compute $$I=\int_0^t \Big[(x^2+x+1)e^{-x}-x^2 \Big]\,dx=4-\frac{t^3}{3}- e^{-t} \left( t^2+3 t+4\right)$$ $t$ being the value of $x$ such that $$(x^2+x+1)e^{-x}=x^2$$

There is a formal definition since we can write $$e^{-x}=\frac {(x-0)(x-0)}{(x-a)(x-b)}$$ where $(a,b)$ are the roots of the quadratic. Then the solution is given in terms of the generalized Lambert function (have a look at equations $(4)$ and $(5)$).

This is formally nice but not very practical.

Knowing (by inspection or graphing) that $t$ is just above $1$, make $$f(x)=(x^2+x+1)e^{-x}-x^2=$$ $$\left(\frac{3}{e}-1\right)-2 (x-1)-\left(1+\frac{1}{2 e}\right) (x-1)^2+\frac 1e\sum_{n=4}^\infty (-1)^n \frac{(n-3) (n-1)}{n!}(x-1)^n$$ Neglecting the summation $$f(x)=\left(\frac{3}{e}-1\right)-2 (x-1)-\left(1+\frac{1}{2 e}\right) (x-1)^2+O\left((x-1)^4\right)$$ and using series reversion, we have the approximation $$x=1+y-\frac{(1+2 e) }{4 e}y^2+\frac{(1+2 e)^2}{8 e^2} y^3+O\left(y^{4}\right)$$ where $y=-\frac{1}{2} \left(f(x)-\frac{3}{e}+1\right)$. Making $f(x)=0$ as desired, then the approximation of $t$.

Using my phone, $t\sim 1.05$ (using a computer $t=1.05033$, the "exact" solution being $t=1.05032$).Still without computer make $t\sim\frac{21}{20}$ to make (still with my phone) $$I \sim \frac{28913}{8000}-\frac{3301}{400e}e^{-1/20}$$

Now, using $e\sim \frac{19}{7}$ and $e^{-1/20}\sim 1-\frac 1 {20}=\frac{19}{20}$ then $$I \sim \frac{28913}{8000}-\frac{3301}{400}\times \frac7{19}\times\frac{19}{20}=\frac{2903}{4000}=0.72575$$

while the "exact" solution (using Newton method to compute $t$) would be $0.72626$. That is to say that we have an absolute relative error equal to $0.07$%.