Question

Need to calculate the Area under the "f(x)=e^-x²" between the points where y=0.5. I found out that the x value here is "±sqrt(ln(2))".
Now I did the following:
Integrate from -sqrt(ln(2)) to sqrt(ln(2)) f(x)=e^-x²
However, If I look up the answer it says:
Integrate from -sqrt(ln(2)) to sqrt(ln(2)) f(x)=e^(-x²)-(1/2)
Why the -(1/2)?

Answer 1

Mine
\[\int\limits_{-\sqrt{\ln 2}}^{\sqrt{\ln 2}}e ^{-x²}\]
The correct one:
\[\int\limits_{-\sqrt{\ln 2}}^{\sqrt{\ln 2}}e ^{-x²}-0.5\]

Answer 2

Okay you are finding the area of the region between y=e^-x^x and y=.5

Answer 3

http://www.wolframalpha.com/input/?i=Plot%5Be^%28-x%29^2%2C0.5%2C {x%2C-2%2C2}%5D

Answer 4

http://www.wolframalpha.com/input/?i=area+between+f%28x%29%3De^-x^2+and+y%3D.5+

Answer 5

Just doing integral of e^-x² find you area of region between the function and y=0 which is not what you want

Answer 6

its - 1/2 because you are calculating the area inside or in between
y=e^-x^2 and y=0.5

Answer 7

Ah, that makes sense. Thanks guys!