Final answer:
To find the area between the curves y=e⁻x and y=e²x/3 from x=0 to x=3, we set the two equations equal to each other, find the points of intersection, and calculate the integral of their difference. The area is 3e⁶ + e⁻³ + 1.
Step-by-step explanation:
To find the area between the curves y=e⁻x and y=e²x/3 from x=0 to x=3, we need to find the integral of the difference between the two curves. First, let's find the points of intersection between the two curves by setting them equal to each other:
e⁻x = e²x/3
To solve for x, we take the natural logarithm of both sides:
-x = 2x/3
Multiplying both sides by -3 and rearranging the equation, we get:
3x + 2x = 0
5x = 0
x = 0
So the curves intersect at x=0. Now, we can find the area by evaluating the integral:
Area = ∫(e²x/3 - e⁻x) dx from 0 to 3
Calculating the integral, we get:
Area = [3e²x - (-e⁻x)] from 0 to 3
Area = 3e²(3) - (-e⁻(3)) - 3e²(0) - (-e⁻(0))
Area = 3e⁶ + e⁻³ + 1
So the area between the curves y=e⁻x and y=e²x/3 from x=0 to x=3 is 3e⁶ + e⁻³ + 1.