What is the area of the region bounded by the graphs y=e^x, y=1, x=-1?
I see a thin "triangular" region in quadrant II
form x = -1 to x=0
the "height" is 1 - e^x
area = ∫(1 - e^x) dx from -1 to 0
= [ x - e^x] from -1 to 0
= 0 - 1 - (-1 - e^-1)
= e^-1
or
= 1/e
To find the area of the region bounded by the graphs y = e^x, y = 1, and x = -1, we can use integration.
Step 1: First, let's plot the given graphs to get a visual representation of the region.
The graph of y = e^x is an exponential curve that starts at (0, 1) and increases rapidly as x increases.
The graph of y = 1 is a horizontal line located at y = 1.
The line x = -1 is a vertical line passing through x = -1.
Step 2: Now, we need to find the points of intersection between these curves.
To find the point(s) of intersection between y = e^x and y = 1, we set the two equations equal to each other and solve for x:
e^x = 1
Taking the natural logarithm (ln) of both sides, we get:
x = ln(1)
Since ln(1) = 0, the two curves intersect at x = 0.
Step 3: Now that we have determined the range of integration, we can calculate the area using definite integration.
Since we are interested in the area bounded by the graphs y = e^x, y = 1, and x = -1, the integral will be:
Area = ∫[from x = -1 to x = 0] (e^x - 1) dx
Now, we integrate the function (e^x - 1) with respect to x from -1 to 0.
Area = ∫(e^x - 1) dx
= [e^x - x] [from x = -1 to x = 0]
= (e^0 - 0) - (e^(-1) - (-1))
= (1 - 0) - (1/e + 1)
Simplifying further, we get:
Area = 1 - 1/e - 1
= -1/e
Therefore, the area of the region bounded by the graphs y = e^x, y = 1, and x = -1 is -1/e.