Calculate the area between curves y=e^x and y=-e^x on the interval [0,1]
what's the problem? e^x is about the simplest thing to integrate
integral[0,1] e^x - (-e^x) dx
= 2e^x [0,1]
= 2(e-1)
that is how I worked it out as well but I did not know if it was that simple I find ways of making simple things hard sometimes thanks for the help
To calculate the area between the curves y = e^x and y = -e^x on the interval [0,1], we can use integration.
Step 1: Find the intersection points.
To find the intersection points between the two curves, we set the two equations equal to each other and solve for x:
e^x = -e^x
Dividing both sides by e^x, we get:
1 = -1
This equation has no real solution, which means that the curves y = e^x and y = -e^x do not intersect on the interval [0,1].
Step 2: Determine the limits of integration.
Since there are no intersection points, we cannot directly calculate the area between the curves. However, we can calculate the integral of the absolute difference between the two functions on the interval [0,1].
The area between the curves y = e^x and y = -e^x on the interval [0,1] can be calculated as:
∫[0,1] |e^x - (-e^x)| dx
Step 3: Integrate the absolute difference.
The absolute difference between the two functions is given by:
|e^x - (-e^x)| = |2e^x|
So, our integral becomes:
∫[0,1] 2e^x dx
Step 4: Evaluate the integral.
Integrating 2e^x with respect to x, we get:
∫[0,1] 2e^x dx = 2[e^x] evaluated from 0 to 1
Plugging in the upper and lower limits of integration, we have:
2[e^1 - e^0] = 2[e - 1]
Therefore, the area between the curves y = e^x and y = -e^x on the interval [0,1] is 2(e-1) square units.