Problem 25 - Calculus for Biology and Medicine

Question

Find the areas of the regions bounded by the lines and curves. \(y=e^{x / 2}, y=-x, x=0, x=2\)

Step-by-Step Solution

Answer

The area is \(2e\) square units.

Step 1: Understand the problem

We need to find the area enclosed by the curves \(y=e^{x/2}\), \(y=-x\), and the lines \(x=0\) and \(x=2\). Start by sketching these functions on a coordinate plane to see where they intersect and form a bounded region.

Step 2: Identify Points of Intersection

Find the points where the curves \(y=e^{x/2}\) and \(y=-x\) intersect by setting them equal: \\(e^{x/2} = -x\). This can be solved graphically or using approximation techniques, as it does not yield a simple algebraic solution. Within the range \(x = 0\) to \(x = 2\), they intersect at \(x = 0\).

Step 3: Set up the integral

The area between two curves is found by integrating the difference of the curves within the bounds of integration. Set up the integral of \(e^{x/2} - (-x)\) from \(x=0\) to \(x=2\). This is the integral \[\int_0^2 (e^{x/2} + x) \, dx\].

Step 4: Compute the integral

Calculate the definite integral: \1. Find the antiderivative of \(e^{x/2}\), which is \(2e^{x/2}\). \2. Find the antiderivative of \(x\), which is \(\frac{x^2}{2}\). \3. Evaluate these from \(x=0\) to \(x=2\): \\[\left[ 2e^{x/2} \right]_0^2 + \left[ \frac{x^2}{2} \right]_0^2\].

Step 5: Solve the definite integral

Insert the limits into the antiderivative expressions and subtract: \\[\left( 2e^{2/2} + \frac{2^2}{2} \right) - \left( 2e^{0/2} + \frac{0^2}{2} \right)\] \\[= (2e + 2) - (2 + 0) \] \\[= 2e + 2 - 2 \] \\[= 2e\].

Step 6: Interpret the result

The bounded area is \(2e\) square units. This is the area between the two curves and the vertical lines from \(x=0\) to \(x=2\).

Key Concepts

Definite Integral

The concept of the definite integral is a fundamental part of calculus, primarily used to calculate areas under curves and between curves. The definite integral

• is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• It measures the accumulated area under a curve from \( x = a \) to \( x = b \).

For the exercise, the definite integral is used to find the area between the given curves \( y = e^{x/2} \) and \( y = -x \) from \( x = 0 \) to \( x = 2 \). To solve the integral, we find the antiderivative of the function. The integral of \( (e^{x/2} + x) \) is calculated using basic integral rules:

• The antiderivative of \( e^{x/2} \) is \( 2e^{x/2} \) because applying the chain rule during differentiation results in the original expression.
• The antiderivative of \( x \) is \( \frac{x^2}{2} \), coming from basic power rule application in integration.

Subsequently, by substituting \( x = 2 \) and \( x = 0 \) into these antiderivatives and subtracting, we find the total area enclosed by these curves. This calculation ultimately gives the numerical answer, \( 2e \), which represents the area between these curves from the start to the end of the specified interval.

Area Between Curves

The concept of finding the area between curves is central in applications of calculus. When curves overlap or create closed regions, the area in between them is often of interest—useful in real-world problems across physics, engineering, and economics.

• To find the area between two curves, we first determine the points of intersection to establish bounds of integration.
• The area is then the integral of the top function minus the bottom function, over the interval bounded by their intersection points.

In this case, the functions are \( y = e^{x/2} \) and \( y = -x \). The integral \( \int_0^2 (e^{x/2} + x) \, dx \) represents this area, calculated by integrating the difference between the functions across the specified range from \( x = 0 \) to \( x = 2 \). The concept boils down to subtracting the lower curve (\( y = -x \) becomes \( +x \)) from the upper curve \( y = e^{x/2} \) and solving through integration to find the area neatly encapsulated between the two.

Points of Intersection

Identifying points of intersection is the first step to finding areas between curves. It tells where the curves meet, serving as integral bounds. Here's how you can approach it:

• Set the two functions equal to each other: for this problem, \( e^{x/2} = -x \).
• Solve this equation for \( x \). Analytical solutions might not always be possible; when complex, use numerical approximation or graphical methods.

In our particular exercise, the functions intersect within the bounds at \( x = 0 \). Because the integral is evaluated from \( x = 0 \) to \( x = 2 \), finding additional intersections wasn't necessary here.Understanding these intersection points ensures the correct evaluation, so we integrate over the appropriate interval. This provides accurate computation of the total area bound between the specified curves and lines.