## Calculus: Early Transcendentals 8th Edition

The statement is False, so $$\int_{-\infty}^{\infty} f(x) d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t}f( x) d x$$
The statement is false, so $$\int_{-\infty}^{\infty} f(x) d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t}f( x) d x$$ Now, we give an example that disproves the statement. Let $f(x)=x$ $$\int_{-\infty}^{\infty} x d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t} xd x$$ (a) The given integral $$\int_{-\infty}^{\infty} x d x$$ is an improper integral, hence $$\int_{-\infty}^{\infty} x d x =\int_{-\infty}^{0} x d x+\int_{0}^{\infty} x d x$$ and $$\begin{split} \int_{0}^{\infty} x d x & = \lim _{t \rightarrow \infty} \int_{0}^{t} x d x\\ & =\lim _{t \rightarrow \infty}\left[\frac{1}{2} x^{2}\right]_{0}^{t} \\ & =\lim _{t \rightarrow \infty}\left[\frac{1}{2} t^{2}-0\right] \\ & = \infty, \end{split}$$ so the integral $$\int_{0}^{\infty} x d x$$ is divergent, and hence, $$\int_{-\infty}^{\infty} x d x$$ is divergent. (b) $$\begin{split} \int_{-t}^{t} x d x & = \left[\frac{1}{2} x^{2}\right]_{-t}^{t} \\ & =\left[\frac{1}{2} t^{2}-\frac{1}{2} t^{2}\right] \\ & = 0, \end{split}$$ so $$\lim _{t \rightarrow \infty} \int_{-t}^{t} x d x=0$$ Therefore, from (a) and (b) we get $$\lim _{t \rightarrow \infty} \int_{-t}^{t} x d x \ne \int_{-\infty}^{\infty} x d x$$