Question

Compute the integral

₁∫[infinity] e⁻ˣ dx. State your answer as a decimal rounded to the thousandths place.

Answer 1

Rounded to the thousandths place, the value of the integral is approximately 0.367.

Step-by-step explanation:

The integral to compute is:

∫₍₁/infinity₎ e^(-x) dx

To calculate this integral, Evaluate the limit of the definite integral as the upper bound approaches infinity. The integral of e^(-x) is known to be -e^(-x), so we have:

lim₍b→infinity₎ [ -e^(-x) ] evaluated from 1 to b

Evaluating the definite integral, we get:

lim₍b→infinity₎ [ -e^(-b) - (-e^(-1)) ]

As the upper bound approaches infinity, the term -e^(-b) approaches zero since e^(-b) decreases exponentially as b becomes larger.

Thus, the integral simplifies to:

- (-e^(-1)) = e^(-1) = 0.367

Answer 2

To solve the integral from 1 to infinity of e^-x dx, find the indefinite integral and apply the limits, yielding a result of approximately 0.368, rounded to the thousandths place.

Step-by-step explanation:

The student has asked to compute the integral from 1 to infinity of e-x dx. To solve this, you can use the fundamental theorem of calculus by finding an indefinite integral, then applying limits. The indefinite integral of e-x dx is -e-x (since the derivative of -e-x is e-x). Applying the limits from 1 to infinity:

∞
∫ e-x dx = -e-x |1∞

We get:

lim (x → ∞) (-e-x) - (-e-1) = 0 - (-e-1) = e-1 ≈ 0.368

Thus, the decimal value of the integral, rounded to the thousandths place, is approximately 0.368.