Find the first quadrant area under y = e -2x
I assume you mean y = e^(-2x)
area = ∫e^(-2x) dx from 0 to ∞
= [ (-1/2)e^(-2x) ] from 0 to ∞
let's look at (-1/2) e(^(-2x)
or -1/(2e^(2x) as x ---> ∞
so as the denominator becomes larger and larger, the quotient becomes smaller and smaller
that is ,
limit -1/(2e^(2x) as x---> ∞ = 0
so area = 0 - (-1/e^0
= 1
let's test our answer, let the area be from 0 to 100
area = -1/e^200 - ( -1/e^0)
= 1.38x10^-87 + 1
= pretty close to 1
To find the area under the curve of the function y = e^(-2x) in the first quadrant, we need to integrate the function over the appropriate interval.
The first quadrant is defined by x ≥ 0 and y ≥ 0.
To find the area, we need to integrate the function with respect to x from x = 0 to the point where the curve intersects the x-axis. To find this point, we set y = 0 and solve for x:
0 = e^(-2x)
Since e^(-2x) is always positive, there are no solutions that satisfy y = 0. This means that the curve does not intersect the x-axis in the first quadrant.
Therefore, the area under the curve y = e^(-2x) in the first quadrant is infinite since it extends indefinitely in the positive y-direction.
To find the first quadrant area under the curve y = e^(-2x), we need to integrate this function over the appropriate interval.
The first quadrant includes positive x-values and positive y-values. Therefore, we need to find the values of x where the curve intersects the x-axis and the y-axis.
To find the x-intercept, we set y = 0:
0 = e^(-2x)
Taking the natural logarithm of both sides:
ln(0) = ln(e^(-2x))
0 = -2x
Now we solve for x:
0 = -2x
x = 0
So, the x-intercept of the curve is at x = 0.
To find the y-intercept, we set x = 0:
y = e^(-2 * 0)
y = e^0
y = 1
So, the y-intercept of the curve is at y = 1.
Now we have the x and y-values that define the first quadrant area: (0, 0) and (0, 1).
To find the area under the curve, we integrate the function y = e^(-2x) with respect to x over the interval [0, ∞).
∫[0, ∞) e^(-2x) dx
Integrating this function gives us:
∫[0, ∞) e^(-2x) dx = -1/2 * e^(-2x) | [0, ∞)
Now we evaluate the definite integral from 0 to ∞:
-1/2 * e^(-2 * ∞) - (-1/2 * e^(-2 * 0))
Since e raised to any negative power approaches zero as x goes to infinity, we can simplify:
-1/2 * 0 - (-1/2 * 1)
= 0 + 1/2
= 1/2
Therefore, the first quadrant area under the curve y = e^(-2x) is 1/2 square units.