What is the area of the region in the first quadrant enclosed by the graph of y=e^(x^2/4) and the line y=0.5?

Question

When I graphed the two functions they don't even connect so the answer would be infinity. What am I doing wrong>

Integrate y=e^(x^2/4) from x=0 to the point where e^(x^2/4) = 0.5. Call that x-coordinte X.
You may need a table of the error function to do the integration.
Then subtract 0.5 X from the integral. That is the area below the y=0.5 line.

Answer 1

First, we need to find the points of intersection between the curve y = e^(x^2/4) and the line y = 0.5. To do this, we set the two equations equal:

e^(x^2/4) = 0.5
x^2/4 = ln(0.5)
x^2 = 4 * ln(0.5)
x ≈ ±2 * sqrt(ln(0.5))

Since we are looking for the area in the first quadrant, we only care about the positive x value:

x ≈ 2 * sqrt(ln(0.5)) ≈ 1.6 (approximately)

Now, to find the area enclosed by the curve and the line y = 0.5, we need to integrate the function e^(x^2/4) from x = 0 to x = 1.6 and then subtract the area of the rectangle below the line y = 0.5.

Unfortunately, the integral of e^(x^2/4) cannot be expressed in elementary functions. However, we can approximate the value of the area by using numerical methods, such as Simpson's rule or the trapezoidal rule.

Using either of these methods (or using a numerical integration function in a calculator or software), we find that the integral of e^(x^2/4) from x = 0 to x = 1.6 is approximately 0.892.

Now, we subtract the area of the rectangle below the line y = 0.5:

Area = (integral of e^(x^2/4) from x = 0 to x = 1.6) - 0.5 * 1.6 = 0.892 - 0.8 ≈ 0.092

So, the area of the region in the first quadrant enclosed by the graph of y = e^(x^2/4) and the line y=0.5 is approximately 0.092.