The Area Between The Curves

What is the area between the curves？

Definition

The Area Between the Curves represents the region enclosed by the graphs of two or more curves in a coordinate plane. It is a concept found in integral calculus that allows for the calculation of the space that lies between the plotted Definitionfunctions, typically on a Cartesian plane. This area can be determined by integrating the difference between the functions over a specific interval.

How to Find the Area:

To calculate the area A between two curves defined by functions $f(x)$ and $g(x)$ over an interval $[a, b]$, where $f(x)$ is always greater than or equal to $g(x)$ in the interval, you use the formula:

$ A = \int_{a}^{b} [f(x) - g(x)] ,dx $

Steps for Calculation:

1. Identify the Functions: Determine which function is on top ($f(x)$) and which is on the bottom ($g(x)$) for the interval $[a, b]$. The top function is the one with higher values for all $x$ in the interval.

2. Find the Intersection Points: If the interval $[a, b]$ isn't given, you'll need to find where the functions intersect to use as the interval boundaries. Solve $f(x) = g(x)$ to find these points.

3. Integrate the Difference: The integral of the difference between the top function and the bottom function over the interval gives the area between the curves.

4. Calculate: Compute the definite integral from step 3 to find the area.

Important Considerations:

• Intersection Points: If $f(x)$ and $g(x)$ intersect within your interval of interest, these intersection points typically split the area into distinct sections that might need to be calculated separately.

• Functions Switching Places: If $f(x)$ and $g(x)$ switch which one is on top within the interval, you'll need to break the integral into parts where each function consistently remains on top within each subinterval.

Solved Examples of Finding the Area Between the Curves

Example 1: Basic Vertical Curves

Objective: Find the area between $f(x) = x^2$ and $g(x) = x + 2$ in the interval where they intersect.

Step 1: Find the Intersection Points

Solve $x^2 = x + 2$

$x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0$

So, $x = 2$ and $x = -1$. These are our limits of integration.

Step 2: Set Up the Integral

Since $g(x)$ is above $f(x)$ between $-1$ and $2$ (you can verify this by plugging in values or graphing), the area $A$ is:

$A = \int_{-1}^{2} [(x + 2) - (x^2)] ,dx$

Step 3: Calculate the Integral

$A = \int_{-1}^{2} (-x^2 + x + 2) ,dx$ $A = \left[ -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x \right]_{-1}^{2}$ $A = \left(-\frac{8}{3} + 2 + 4\right) - \left(\frac{1}{3} - \frac{1}{2} - 2\right)$ $A = -\frac{8}{3} + \frac{3}{2} + 2$

$A = 9/2$

Conclusion for Example 1:

The area between the curves $y = x^2$ and $y = x + 2$ from $x = -1$ to $x = 2$ is $9/2$ square units.

Example 2: Area Between a Parabola and a Line

Objective: Calculate the area between $y = 4 - x^2$ and $y = 2x$ where they intersect.

Step 1: Find Intersection Points

Solve $4 - x^2 = 2x$

$x^2 + 2x - 4 = 0$

Using the quadratic formula, we find:

$x = \frac{-2 \pm \sqrt{4 + 16}}{2} = -1 \pm \sqrt{5}$

So, the intersection points are $-1 + \sqrt{5}$ and $-1 - \sqrt{5}$.

Step 2: Set Up the Integral

Here, $4 - x^2$ is above $2x$ between the intersection points. The area is:

$A = \int_{-1 - \sqrt{5}}^{-1 + \sqrt{5}} [(4 - x^2) - (2x)] ,dx$

Step 3: Calculate the Integral

$A = \int_{-1 - \sqrt{5}}^{-1 + \sqrt{5}} (-x^2 - 2x + 4) ,dx$

$A = \left[ -\frac{1}{3}x^3 - x^2 + 4x \right]_{-1-\sqrt{5}}^{-1+\sqrt{5}}$

Substituting the limits into the antiderivative yields a more involved computation, but after evaluating, you would get the numeric result representing the area between these curves over the given interval.

Example 3: Area between $y = x^2$ and $y = 2x + 3$

Step 1: Intersection Points Solve $x^2 = 2x + 3$: $x^2 - 2x - 3 = 0 \rightarrow (x-3)(x+1) = 0$ $x = 3, , x = -1$

Step 2: Integral Setup $A = \int_{-1}^{3} ((2x + 3) - x^2) , dx$

Step 3: Calculate the Area $A = \left[ x^2 + 3x - \frac{x^3}{3} \right]_{-1}^{3}$ $A = \left( 9 + 9 - \frac{27}{3} \right) - \left( 1 + (-3) - \frac{-1}{3} \right)$ $A = 12 - (-2 +\frac{1}{3}) = 12 + 2 - \frac{1}{3} = 14 - \frac{1}{3} = \frac{41}{3}$

Conclusion:

The area between the curves from $x = -1$ to $x = 3$ is $\frac{41}{3}$ square units.

Example 4: Area between $y = \sqrt{x}$ and $y = x/2$

Step 1: Intersection Points Solve $\sqrt{x} = x/2$: $x = \frac{x^2}{4} \rightarrow 4x = x^2 \rightarrow x^2 - 4x = 0$ $x(x - 4) = 0$ $x = 0, , x = 4$

Step 2: Integral Setup $A = \int_{0}^{4} (\sqrt{x} - \frac{x}{2}) , dx$

Step 3: Calculate the Area $A = \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{4} \right]_{0}^{4}$ $A = \left( \frac{32}{3} - 4 \right) - 0 = \frac{32}{3} - \frac{12}{3} = \frac{20}{3}$

Conclusion:

The area between the curves from $x = 0$ to $x = 4$ is $\frac{20}{3}$ square units.

Example 5: Area between $y = \cos(x)$ and $y = \sin(x)$ from $x = 0$ to $x = \frac{\pi}{2}$

Step 1: Intersection Points Solve $\cos(x) = \sin(x)$: $x = \frac{\pi}{4}$ (primary intersection within the interval)

Step 2: Integral Setup Since $\sin(x) \geq \cos(x)$ in $[0, \frac{\pi}{4}]$ and vice versa in $[\frac{\pi}{4}, \frac{\pi}{2}]$: $A = \int_{0}^{\frac{\pi}{4}} (\sin(x) - \cos(x)) , dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos(x) - \sin(x)) , dx$

Step 3: Calculate the Area

$A = \left[ -\cos(x) - \sin(x) \right]*{0}^{\frac{\pi}{4}} + \left[ \sin(x) + \cos(x) \right]\cdot {\frac{\pi}{4}}^{\frac{\pi}{2}}$ $A = (-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}) - (-1 - 0) + (1 + 0) - (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2})$

$A = 2 - \sqrt{2}$

Conclusion:

The area between $\cos(x)$ and $\sin(x)$ from $x = 0$ to $x = \frac{\pi}{2}$ is $2 - \sqrt{2}$ square units.

Example 6: Area under $y = e^{-x}$ from $x = 1$ to $x = 2$

Step 1: Since this involves area under one curve, consider the "other" curve as $y = 0$.

Step 2: Integral Setup $A = \int_{1}^{2} e^{-x} , dx$

Step 3: Calculate the Area $A = \left[ -e^{-x} \right]_{1}^{2} = -e^{-2} - (-e^{-1}) = e^{-1} - e^{-2}$

Conclusion:

The area under $y = e^{-x}$ from $x = 1$ to $x = 2$ is $e^{-1} - e^{-2}$ square units.

Example 7: Symmetrical Area around the y-axis between $y = x^3$ and $y = x$

Step 1: Since symmetry is mentioned, calculate for $x \geq 0$ and double it.

Step 2: Intersection Points Solve $x^3 = x$: $x^3 - x = 0 \rightarrow x(x^2 - 1) = 0 \rightarrow x(x + 1)(x - 1) = 0$ $x = 0, \pm 1$

Step 3: Integral Setup for the positive side (and then double it) $A = 2\int_{0}^{1} (x - x^3) , dx$

Step 4: Calculate the Area $A = 2\left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = 2\left( \frac{1}{2} - \frac{1}{4} \right) = 2\left( \frac{1}{4} \right) = \frac{1}{2}$

Conclusion:

The symmetrical area around the y-axis between $y = x^3$ and $y = x$ is $\frac{1}{2}$ square units.

FAQs about The area Between the Curves

Q: What is area under the curve called?

A: The area under a curve, in the context of mathematics, particularly calculus, is referred to as the "integral" of the function defining the curve over a specified interval. When computing this area, especially when the function $f(x)$ is plotted on a Cartesian plane with the x-axis serving as the baseline, the process involves finding the definite integral of $f(x)$ from a point $a$ to $b$, symbolically represented as:

$ \int_{a}^{b} f(x) , dx $

This definite integral quantifies the total area bounded by the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$, considering areas above the x-axis as positive and areas below the x-axis as negative.

Q: How do you find the area between three curves?

A: Finding the area between three curves involves a few more steps compared to finding the area between two curves, primarily because we need to consider how these curves intersect and enclose a region. Here’s a generalized stepwise approach:

Step 1: Sketch the Curves

Draw a rough sketch of the curves $f(x)$, $g(x)$, and $h(x)$ to understand their relative positions and identify the enclosed region(s) for which you’re calculating the area.

Step 2: Identify Intersection Points

Find the points where these curves intersect one another. This typically involves solving the equations $f(x) = g(x)$, $f(x) = h(x)$, and $g(x) = h(x)$ to find the x-coordinates of the intersection points. These points define the limits of integration.

Step 3: Determine the Enclosed Region

Based on the sketch and intersection points, determine the region(s) enclosed by the curves. Sometimes, there might be multiple distinct areas to consider, especially if the curves intersect at more than two points.

Step 4: Set Up the Integral(s)

For each enclosed area, you'll need to set up an integral or a series of integrals. The typical form is:

$ \text{Area} = \int_{a}^{b} \big|f(x) - g(x)\big|,dx + \int_{c}^{d} \big|h(x) - g(x)\big|,dx $

Or any variation that accounts for the specific curves enclosing the area. Here, $a$, $b$, $c$, and $d$ are the x-coordinates of the intersection points, and you choose $f(x)$, $g(x)$, and $h(x)$ based on which curve is uppermost or lowermost in the region being integrated.

Step 5: Compute the Integral(s)

Calculate the integral(s) using appropriate techniques. For definite integrals, this will involve finding the antiderivatives and evaluating them at the upper and lower limits of integration.

Step 6: Add the Areas

If the region of interest is composed of multiple areas (from separate integrals), sum them to get the total area enclosed by the three curves.

Example

Suppose you have curves defined by $y = f(x) = x^2$, $y = g(x) = 2x+1$, and a horizontal line $y = h(x) = 3$, and you want to find the area enclosed by these curves.

1. Sketch and Identify Intersection Points: Let's say $f(x)$ and $g(x)$ intersect at $x = -1$ and $x = 2$, and $g(x)$ intersects $h(x)$ at some points $c$ and $d$.

2. Determine Enclosed Region: Assume $f(x)$ lies below $g(x)$ and $h(x)$ above them in a certain interval.

3. Set Up Integral(s):

   • For $f(x)$ and $g(x)$, between $-1$ and $2$: $\int_{-1}^{2} (g(x) - f(x)),dx$.
   • For $g(x)$ and $h(x)$, between points $c$ and $d$: $\int_{c}^{d} (h(x) - g(x)),dx$.

4. Compute the Integral(s): Solve each integral based on the given functions.

5. Sum the Areas: Add the results to get the total area enclosed by the three curves.

Q: What is the formula for the distance between two curves?

A: The formula to calculate the distance between two curves depends on the context and the orientation of the curves. Generally, the distance between two curves $y = f(x)$ and $y = g(x)$ in a given interval $[a, b]$ on the $x$-axis can be defined as the absolute difference between the functions, integrated over that interval. However, for a more straightforward, point-to-point comparison, the vertical or horizontal distance at a specific $x$ or $y$ value can be found using simpler expressions.

Vertical Distance Between Two Curves

For any given value of $x$, if you want to find the vertical distance between $y = f(x)$ and $y = g(x)$ where $f(x)$ is above $g(x)$, the distance $D$ at a point $x$ can be calculated as:

$D = |f(x) - g(x)|$

This gives the vertical distance between the two curves at that specific point.

Horizontal Distance Between Two Curves

For curves defined as $x = f(y)$ and $x = g(y)$, the horizontal distance between them at a given point $y$ is:

$D = |f(y) - g(y)|$

Average Distance

The average distance between two curves $f(x)$ and $g(x)$ over an interval $[a, b]$ can be expressed as:

$D_{\text{average}} = \frac{1}{b-a}\int_{a}^{b} |f(x) - g(x)| , dx$

Example: Vertical Distance Between Two Functions

Given two functions $f(x) = x^2$ and $g(x) = 2x + 3$, to find the vertical distance between them at $x = 1$:

$D = |f(1) - g(1)| = |1^2 - (2 \cdot 1 + 3)| = |1 - 5| = 4$