Question #186250

Improper Integral (Integrals with Infinite Limits)


∫e^x dx from - ∞ to 1


1
Expert's answer
2021-05-07T10:23:17-0400

Given Improper integral is-


I=1exdxI=\int_{-\infty}^1e^xdx


So, This integral can be written as-

I=limpp1exdx=limpexp1=limp(eep)I=lim_{p\to \infty}\int_-p^1e^xdx \\ =lim_{p\to \infty}e^x|_p^1 \\ =lim_{p\to \infty} (e-e^{-p})\\


=e1e=e1=e0=e=e-\dfrac{1}{e^{\infty}} =e-\dfrac{1}{\infty}=e-0=e



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