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.