Problem 11

Question

Find where the curves in $1-12$ intersect, draw rough graphs, and compute the area between them. $$ y=e^{x} \text { and } y=e^{2 x-1} \text { and } x=0 $$

Step-by-Step Solution

Verified

Answer

The curves intersect at (1, e), and the area between them is $\frac{1}{2}e - 1 + \frac{1}{2e}$.

1Step 1: Find the Intersection Points

To find where the curves intersect, equate the equations of the curves: \[ e^x = e^{2x-1} \]Taking natural log (\ln) on both sides leads to:\[ x = 2x - 1 \]Re-arranging gives:\[ 1 = x \]Therefore, the curves intersect at the point $(1, e)$. We must also consider $x=0$, which gives another point of $(0, 1)$ since $e^x|_{x=0} = 1$ and $e^{2x-1}|_{x=0} = e^{-1} = \frac{1}{e}$. However, since $e^{2 imes 0 - 1} ≠ e^0$, only $(0, 1)$ is relevant from the boundary.

2Step 2: Rough Graph of the Curves

Sketch a rough graph of both curves and the vertical line $x = 0$. The graph $y=e^x$ is an exponential curve starting at $y=1$ when $x=0$ and increasing. The graph $y=e^{2x-1}$ starts below at $x=0$ due to $e^{-1}$ and rises faster but intersects $y=e^x$ at the point $(1, e)$. The region of intersection spans from $x = 0$ to $x = 1$.

3Step 3: Set Up the Integral for the Area Between Curves

To find the area between two curves, subtract the lower function from the upper function:\[ A = \int_{0}^{1} (e^x - e^{2x-1}) \, dx\]

4Step 4: Calculate the Integral

Integrate the expression:\[ A = \int_{0}^{1} e^x \, dx - \int_{0}^{1} e^{2x-1} \, dx\]The integral of $e^x$ is $e^x$, and the integral of $e^{2x-1}$ is $\frac{1}{2} e^{2x-1}$.Evaluating from $x = 0$ to $x = 1$:\[ A = [e^x]_{0}^{1} - \left[\frac{1}{2}e^{2x-1}\right]_{0}^{1}\]\[ A = (e^1 - e^0) - \frac{1}{2}(e^{2 \times 1 - 1} - e^{2 \times 0 - 1})\]\[ A = (e - 1) - \frac{1}{2}(e - \frac{1}{e})\]Simplify the expression further:\[ A = (e - 1) - \frac{1}{2}e + \frac{1}{2e}\]Combining like terms gives:\[ A = \frac{1}{2}e - 1 + \frac{1}{2e}\]

5Step 5: Finalize the Answer

Simplify the expression to find the numerical approximation if desired, but the exact area between the curves is: \[ A = \frac{1}{2}e - 1 + \frac{1}{2e}\] This completes the area calculation.

Key Concepts

Exponential FunctionsArea Between CurvesIntegral Calculus

Exponential Functions

Exponential functions form an important class of functions in mathematics. They are characterized by the constant rate of growth or decay that they exhibit.

The general form of an exponential function is given by $ y = a e^{bx} $, where $ e \approx 2.718 $ is the base of the natural logarithm.
In the exercise, the functions given are $ y = e^x $ and $ y = e^{2x-1} $.
An exponential increase occurs when the exponent value is positive. Whereas, if the exponent is negative, the function describes exponential decay.

In the exercise, the function $ y = e^x $ increases steadily as $ x $ increases, starting from $ y = 1 $ when $ x = 0 $. The function $ y = e^{2x-1} $ has a steeper slope, as indicated by the factor $ 2 $ representing faster growth. However, it starts below the function $ y = e^x $ at $ x=0 $ due to the $-1$ shift, represented by starting at $ e^{-1} $. Understanding these curves' behavior is crucial for identifying where they intersect and the shape of the area they enclose.

Area Between Curves

Finding the area between two curves involves determining the region that is enclosed by these curves. This topic is common when working with continuous functions in calculus.

When curves intersect, they form enclosed areas that can be calculated using integral calculus.
In the exercise, the two functions intersect at the point $(1, e)$ and between $x = 0$ and $x = 1$.
To compute the area between these curves $ y = f(x) $ and $ y = g(x) $ over an interval $[a, b]$, set up the integral as $ \int_{a}^{b} |f(x) - g(x)| \, dx $.

For this problem, $ f(x) = e^x $ is always greater than $ g(x) = e^{2x-1} $ within the interval $ [0,1] $. The absolute value is redundant in this specific case since $ e^x $ is consistently on top. Calculating the area thus involves integrating $ e^x - e^{2x-1} $ from $ 0 $ to $ 1 $. This integral tells us how the distance between these curves accumulates over the given range.

Integral Calculus

Integral calculus is a powerful branch of mathematics primarily concerned with finding areas under curves and between them. It extends to provide solutions for various real-world applications.

Integrals can be understood as the inverse operation of derivatives, providing accumulation of quantities.
Definite integrals are particularly useful in finding exact areas, such as the area between curves.
In the exercise, the integral $ \int_{0}^{1} (e^x - e^{2x-1}) \, dx $ is evaluated.

To calculate the integral of an exponential function, remember that $ \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C $, where $ C $ is the constant of integration.
Given the functions within the exercise:
- The integral of $ e^x $ is $ e^x $, and the indefinite integral thus becomes $ [e^x]_{0}^{1} = e^1 - e^0 $.
- For $ e^{2x-1} $, a substitution is helpful, leading to the integral being $ \frac{1}{2} e^{2x-1} $, evaluated from 0 to 1.
The calculated difference between these integrals provides the area, resulting in an exact value: $ \frac{1}{2} e - 1 + \frac{1}{2e} $.
This process exemplifies how integral calculus can be systematically applied to yield meaningful results and insights.

Problem 11

Problem 12

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Problem 12

What integral gives the length of Archimedes' spiral $x=t \cos t, y=t \sin t ?$

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Problem 12

Find where the curves in $1-12$ intersect, draw rough graphs, and compute the area between them. $$ y=e \text { and } y=e^{x} \text { and } y=e^{-x} $$

View solution