Question

The value of the integral \( \int_0^\infty e^{-y} \, dy \) is equal to ......... (round off to 2 decimal places)

Hint:

The integral \( \int_0^\infty e^{-y} \, dy \) is a standard integral that evaluates to 1, as it represents the area under the exponential decay curve.

Answer 1

Correct Answer: 0.99

Step 1: Recognize the integral.
This is a standard exponential decay integral: \[ \int_0^\infty e^{-y} \, dy \] We know that the integral of \( e^{-y} \) is \( -e^{-y} \).
Step 2: Evaluate the integral.
Now, evaluate the definite integral: \[ \int_0^\infty e^{-y} \, dy = \left[ -e^{-y} \right]_0^\infty \] At \( y = \infty \), \( e^{-\infty} = 0 \), and at \( y = 0 \), \( e^{0} = 1 \). Therefore: \[ \left[ -e^{-y} \right]_0^\infty = 0 - (-1) = 1 \]
Step 3: Conclusion.
Thus, the value of the integral is \( \boxed{1.00} \) (rounded to two decimal places).