University Calculus: Early Transcendentals (3rd Edition)

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Re-arrange the given integral as follows: $\int_{-\infty}^{0} 2x e^{-x^2} dx+\int_{0}^{\infty} 2x e^{-x^2} dx=\lim\limits_{p \to -\infty}\int_{p}^{0} 2x e^{-x^2} dx+\lim\limits_{q \to \infty}\int_{0}^{q} 2x e^{-x^2} dx$ ....(1) Plug $u(x)=x^2 \implies du=2x dx$ and $u(p)=p^2; u(q)=q^2$ Then, we have $\int_{p^2}^{1} e^{-u} du+\int_{1}^{q^2} e^{-u} du=-1+e^{-p^2}+(-e^{-q^2}+1)$ Thus, equation (1) becomes $\lim\limits_{p \to -\infty}(-1+e^{-p^2})+\lim\limits_{q \to \infty}(-e^{-q^2}+1)=-1+1=0$