The area of the region enclosed by the curves y=ex, y=∣ex−1∣, and the y -axis is:
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Step-by-Step Solution
Step 1: Analyze the curves and region
We need to find the area enclosed by three curves: y=ex, y=∣ex−1∣, and the y -axis (x=0). First, let's understand the behavior of the function y=∣ex−1∣.
Step 2: Define the absolute value function
The absolute value function y=∣ex−1∣ can be split into two cases. When ex−1≥0, which means x≥0, the function is y=ex−1. When ex−1<0, which means x<0, the function is y=−(ex−1)=1−ex.
Step 3: Identify intersection points
We need to find the intersection points of the curves. The y -axis is x=0. At x=0, y=e0=1 and y=∣e0−1∣=∣1−1∣=0. The curves y=ex and y=1−ex intersect when ex=1−ex, which gives 2ex=1, so ex=1/2. This means x=ln(1/2)=−ln2. At this point, y=1/2. The curves y=ex and y=ex−1do not intersect.
Step 4: Set up the integral for the area
The region is bounded by x=0 on the right. The intersection point of y=ex and y=1−ex is at x=−ln2. For x<0, ex>1−ex. For x>0, ex>ex−1. The area is split into two parts: from x=−ln2 to x=0. In the interval [−ln2,0], the upper curve is y=ex and the lower curve is y=1−ex. For x≥0, the curve y=∣ex−1∣ becomes y=ex−1. The region is bounded by y=ex, y=ex−1 and x=0. However, y=ex is always above y=ex−1, and the y -axis is x=0. The area for x≥0 is not enclosed by the three given curves. Thus, the area is only from x=−ln2 to x=0.
Step 5: Evaluate the integral
Now we evaluate the definite integral. We integrate (2ex−1) from x=−ln2 to x=0. The antiderivative of 2ex−1 is 2ex−x. Plugging in the limits, we get (2e0−0)−(2e−ln2−(−ln2)). Since e0=1 and e−ln2=eln(1/2)=1/2, the expression simplifies to (2−0)−(2⋅1/2+ln2)=2−(1+ln2)=1−ln2.