Calculate the area under the graph of y = e^-x above the x-axis for the interval [1, ∞]

a) 0
b) e^-1 ------> my answer. Can you check for me, please?
c) -e^-1
d) infinity

area= int e^-x from 1 to inf

= -e^-x over limits= -e^-inf + e^-1
your are right.

To calculate the area under the graph of y = e^-x above the x-axis for the interval [1, ∞], we need to find the definite integral of the function over that interval.

The integral of e^-x with respect to x is given by:
∫(e^-x) dx

To evaluate this integral, we can use the power rule of integration.

First, we rewrite the integral as:
∫(e^-x) dx = -∫(e^-x) d(-x)

Then, using the power rule, the integral becomes:
∫(e^-x) dx = -e^-x + C

Next, we evaluate the integral over the specified interval [1, ∞]:

∫[1, ∞](e^-x) dx = [-e^-x] evaluated from 1 to ∞

Now, we substitute ∞ into the expression:
[-e^-∞] - [-e^-1]

Since e^-∞ approaches 0 as x approaches infinity:
0 - (-e^-1) = e^-1

Therefore, the correct answer is b) e^-1.

To calculate the area under the graph of y = e^-x above the x-axis for the interval [1, ∞], we can use the definite integral.

The definite integral of e^-x with respect to x from 1 to infinity is:

∫[1,∞] e^-x dx

To evaluate this integral, we can use the formula for the integral of e^(-kx) with respect to x, which is -1/k * e^(-kx) + C.

Applying this formula to our integral, we get:

∫[1,∞] e^-x dx = -1/1 * e^(-1 * x) | [1,∞]

Evaluating this integral from 1 to infinity, we get:

= (-1 * e^(-1 * ∞)) - (-1 * e^(-1 * 1))

Since e^(-∞) approaches 0, we have:

= (-1 * 0) - (-1 * e^-1)

= 0 + e^-1

= e^-1

So, the correct answer is b) e^-1.