# Let R be the region in the first quadrant enclosed by the graphs of #y=e^(-x^2)#, #y=1-cosx#, and the y axis, how do you find the area of the region R?

I graphed the two functions and also roughly highlighted the area we are trying to calculate:

Overview: Find the integral from zero to first intersection point of the two functions of the upper function minus the lower function

To find intersection point: set two equations equal to each other and solve using calculator ():

(On TI 83 to TI 84+CE: either use Numerical solver under MATH menu, or graph the functions and use intersect under CALC menu)

Area of

Plug this into graphing calculator to get:

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To find the area of the region (R), you need to integrate the difference between the functions (e^{-x^2}) and (1-\cos(x)) with respect to (x) from the point where they intersect to the point where (e^{-x^2}) intersects the y-axis.

- Set the two functions equal to each other and solve for (x) to find the x-coordinate(s) of intersection.
- Integrate (e^{-x^2} - (1 - \cos(x))) with respect to (x) from the lower x-coordinate of intersection to the upper x-coordinate of intersection.
- The result of the integration gives you the area of the region (R).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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