Question Number 4511 by madscientist last updated on 04/Feb/16
$${is}\:{this}\:{true}? \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}\:} {dt}=\mathrm{1}\: \\ $$$${if}\:{so}\:{how}? \\ $$$$ \\ $$
Commented by FilupSmith last updated on 04/Feb/16
![∫e^(f(x)) dx=(1/(f ′(x)))e^(f(x)) +c ∫_0 ^( ∞) e^(−t) dt = −[e^(−t) ]_0 ^∞ =−[e^(−∞) −e^0 ] =−[(1/e^∞ )−e^0 ] =−[0−1] =1 ∴∫_0 ^( ∞) e^(−t) dt=1](https://www.tinkutara.com/question/Q4512.png)
$$\int{e}^{{f}\left({x}\right)} {dx}=\frac{\mathrm{1}}{{f}\:'\left({x}\right)}{e}^{{f}\left({x}\right)} +{c} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} {dt}\:=\:−\left[{e}^{−{t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=−\left[{e}^{−\infty} −{e}^{\mathrm{0}} \right] \\ $$$$=−\left[\frac{\mathrm{1}}{{e}^{\infty} }−{e}^{\mathrm{0}} \right] \\ $$$$=−\left[\mathrm{0}−\mathrm{1}\right] \\ $$$$=\mathrm{1} \\ $$$$ \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} {dt}=\mathrm{1} \\ $$
Commented by prakash jain last updated on 04/Feb/16
$$\int{e}^{−{t}} \:\mathrm{d}{t}=−{e}^{−{t}} \:\mathrm{is}\:\mathrm{correct}. \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{formula}\:\int{e}^{{f}\left({x}\right)} \mathrm{d}{x}=\frac{\mathrm{1}}{{f}\:'\left({x}\right)}{e}^{{f}\left({x}\right)} +{c} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{correct}\:\mathrm{in}\:\mathrm{general}. \\ $$$$\mathrm{For}\:\mathrm{example}\:\int{e}^{−{x}^{\mathrm{2}} } \mathrm{d}{x}=? \\ $$
Commented by FilupSmith last updated on 05/Feb/16
$$\mathrm{I}\:\mathrm{was}\:\mathrm{mainly}\:\mathrm{referring}\:\mathrm{to}\:\mathrm{a}\:\mathrm{function}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{power}\:\mathrm{of}\:\mathrm{one}.\:\mathrm{That}\:\mathrm{is}: \\ $$$${f}\left({x}\right)={ax}+{b} \\ $$