We recognize that the integral shown below is unsolvable using any of our known integration techniques.-11e-x2dxHowever, what ifwe instead rewrote e-x2as a Taylor series and took the definite integral of that?Let us begin by making the second degree Taylor Polynomial, T2(x), for e-x2 while centering ourselves at the point x=0.T2(x)=,(x-,)0+,(x-,)1+,(x-,)2When integrating, we get an approximation that is close to the actual area:-11T2(x)dx=Using graphing software like desmos, compare this against the area under the curve for e-x2on the interval -1,1.Is|an over orunder approximation of the true value?Let us now enhance our Taylor polynomial tobeT4(x), still centered at the point x=0. Update your initial answer using the space below.T4(x)=,(x-,)0+,(x-,)1+,(x-,)2+,(x-,)3+,(x-,)4When integrating, we get an approximation that is close to the actual area:-11T4(x)dx=Again using graphing software, compare this against the area under the curve for e-x2on the interval -1,1.Isan over or underapproximation of the true value?