Answer:
[tex]\int_{1}^{\infty} 6e^{-x} dx= -6 [0 -\frac{1}{e}] =\frac{6}{e}[/tex]
Because [tex]\lim_{x\to\infty} \frac{1}{e^x} = 0[/tex]
So then this integral converges and the value obtained is [tex] \frac{6}{e}[/tex]
Step-by-step explanation:
For this case we want to find the following integral:
[tex] \int_{1}^{\infty} 6e^{-x} dx[/tex]
We can take the 6 out of the integral like this:
[tex] 6 \int_{1}^{\infty} e^{-x}[/tex]
And if we solve the integral we got:
[tex]\int_{1}^{\infty} 6e^{-x} dx= -6 e^{-x} \Big|_1^{\infty}[/tex]
[tex]\int_{1}^{\infty} 6e^{-x} dx= -6 [\lim_{x\to\infty} \frac{1}{e^x} - e^{-1}][/tex]
[tex]\int_{1}^{\infty} 6e^{-x} dx= -6 [0 -\frac{1}{e}]=\frac{6}{e}[/tex]
Because [tex]\lim_{x\to\infty} \frac{1}{e^x} = 0[/tex]
So then this integral converges and the value obtained is [tex] \frac{6}{e}[/tex]
