Sketch the region bounded by f ( x ) = − ( x − 2 ) 2 + 5 \textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5} & g ( x ) = 4 x \textcolor{orange}{g\left(x\right)=4x} on the interval [ 0 , 1 ] \left\lbrack0,1\right\rbrack . Then set up an integral to represent the region's area and evaluate.

Integral = ∫ 0 1 − x 2 + 1 d x \int_0^1-x^2+1dx ; Area = 0.67 0.67

Integral = ∫ 0 1 ( − x 2 + 9 ) d x \int_0^1\left(-x^2+9\right)dx ; Area: 0.33 0.33

Integral = ; ∫ 0 4 ( − x 2 + 9 ) d x \int_0^4\left(-x^2+9\right)dx Area: 21.33 21.33

Integral = ∫ 0 1 − x 2 + 1 d x \int_0^1-x^2+1dx ∫ 0 1 − x 2 + 1 d x ; Area = 22.67 22.67

Shade the region bounded by f ( x ) = sin ⁡ x \textcolor{blue}{f\left(x\right)=\sin x} & g ( x ) = cos ⁡ 2 x \textcolor{orange}{g\left(x\right)=\cos2x} on the interval [ π 6 , 5 π 6 ] \left\lbrack\frac{\pi}{6},\frac{5\pi}{6}\right\rbrack . Then set up an integral to represent the region's area.

A = ∫ π 6 5 π 6 ( sin ⁡ x − cos ⁡ 2 x ) d x A=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(\sin x-\cos2x\right)dx

A = ∫ π 6 5 π 6 ( cos ⁡ 2 x − sin ⁡ x ) d x A=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(\cos2x-\sin x\right)dx

A = ∫ π 6 5 π 6 ( sin ⁡ x − cos ⁡ 2 x ) d x A=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(\sin x-\cos2x\right)dx

9. Graphical Applications of Integrals

Area Between Curves

9. Graphical Applications of Integrals

Area Between Curves: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

To find the area between two curves, set up a definite integral from the lower to the upper bounds, which are determined by the intersection points of the functions. The area can be calculated by integrating the top function minus the bottom function. If the functions intersect within the interval, split the integral into two parts, adjusting the top and bottom functions accordingly. This method ensures accurate area calculations, even when functions switch positions across the interval.

concept

Finding Area Between Curves on a Given Interval

Video duration:

Finding Area Between Curves on a Given Interval Video Summary

To find the area between two curves, we utilize the concept of definite integrals, which represent the area under a curve. When tasked with calculating the area between two functions, $ f(x) $ and $ g(x) $, over a specific interval, the process begins with graphing the functions and shading the area of interest. This visual representation helps clarify the relationship between the curves and the area to be calculated.

For example, consider the functions $ f(x) $ and $ g(x) $ defined on the interval from $ x = 0 $ to $ x = 1 $. To find the area between these curves, we first determine which function is on top and which is on the bottom within the specified interval. The area between the curves can be calculated by integrating the top function and subtracting the integral of the bottom function:

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

This can be combined into a single integral:

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

In our example, if $ f(x) = 4 - x^2 $ and $ g(x) = 2x + 1 $, we set up the integral from $ 0 $ to $ 1 $:

\[ \text{Area} = \int_{0}^{1} ((4 - x^2) - (2x + 1)) \, dx \]

After simplifying the integrand, we have:

\[ \text{Area} = \int_{0}^{1} (-x^2 - 2x + 3) \, dx \]

Next, we apply the power rule to find the antiderivative:

\[ \text{Antiderivative} = -\frac{1}{3}x^3 - x^2 + 3x \]

We evaluate this from $ 0 $ to $ 1 $:

\[ \text{Area} = \left[-\frac{1}{3}(1)^3 - (1)^2 + 3(1)\right] - \left[-\frac{1}{3}(0)^3 - (0)^2 + 3(0)\right] \]

Calculating this gives:

\[ \text{Area} = \left[-\frac{1}{3} - 1 + 3\right] - 0 = \frac{5}{3} \]

This result, $ \frac{5}{3} $, represents the area between the two curves over the interval from $ 0 $ to $ 1 $. It is important to note that this method is valid even if the functions are located below the x-axis, ensuring accurate area calculations regardless of the curves' positions.

As you continue to practice finding areas between curves, remember to visualize the functions, set up the integrals correctly, and apply the fundamental theorem of calculus to evaluate your results.

Problem

Calculate the area of the shaded region between the 2 functions from $x equals 0$ to $x equals 9$

4.5

445

Problem

Calculate the area of the shaded region between $f (x)$ & $g (x)$ contained between $x = - 4$ & $x = - 2$ .

8.17

4.17

-12.17

0.50

Problem

Sketch the region bounded by $f (x) = - {(x - 2)}^{2} + 5$ & $g of x equals 4 x$ on the interval $open bracket 0 comma 1 close bracket$ . Then set up an integral to represent the region's area and evaluate.

Graph showing the area between two curves, f(x) in blue and g(x) in orange, highlighted in yellow.

Integral = $\int_{0}^{1} (- x^{2} + 9) d x$ ; Area: $0.33$

Graph showing two curves, f(x) in blue and g(x) in orange, with the area between them shaded in yellow.

Integral = $\int_{0}^{1} - x^{2} + 1 d x$ ; Area = $0.67$

Graph showing the area between two curves, f(x) in blue and g(x) in orange, highlighted in yellow on the interval [0,1].

Integral = ; $\int_{0}^{4} (- x^{2} + 9) d x$ Area: $21.33$

Graph showing two curves, f(x) in blue and g(x) in orange, with the area between them shaded in yellow.

Integral = $\int_0^1-x^2+1dx$ ; Area = $22.67$

Problem

Shade the region bounded by $f (x) = \sin x$ & $g of x equals cosine 2 x$ on the interval $[\frac{π}{6}, \frac{5 π}{6}]$ . Then set up an integral to represent the region's area.

Graph showing the area between the curves f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\cos 2 x - \sin x) d x$

Graph showing the area between the curves f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\cos 2 x - \sin x) d x$

Graph showing the area between the curves of f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\sin x - \cos 2 x) d x$

Graph showing the area between the curves f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\sin x - \cos 2 x) d x$

example

Finding Area Between Curves on a Given Interval

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Finding Area Between Curves on a Given Interval Video Summary

To find the area between the two functions $ f(x) = |x| + 2 $ and $ g(x) = x^2 $ over the interval from $ x = 0 $ to $ x = 2 $, we first identify which function is on top and which is on the bottom. In this case, $ f(x) $ is above $ g(x) $ within the specified interval. The area can be calculated using a definite integral of the top function minus the bottom function.

We set up the integral as follows:

\[\text{Area} = \int_{0}^{2} (f(x) - g(x)) \, dx = \int_{0}^{2} \left( (|x| + 2) - x^2 \right) \, dx\]

Since we are only considering positive values of $ x $ in the interval from $ 0 $ to $ 2 $, we can simplify $ |x| $ to just $ x $. Thus, the integral becomes:

\[\int_{0}^{2} (x + 2 - x^2) \, dx\]

Next, we rewrite the integrand in standard polynomial form:

\[\int_{0}^{2} (-x^2 + x + 2) \, dx\]

Now, we find the antiderivative:

\[\text{Antiderivative} = -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x\]

We will evaluate this antiderivative from $ 0 $ to $ 2 $ using the Fundamental Theorem of Calculus:

\[\text{Area} = \left[ -\frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 + 2(2) \right] - \left[ -\frac{1}{3}(0)^3 + \frac{1}{2}(0)^2 + 2(0) \right]\]

Calculating the upper bound:

\[-\frac{1}{3}(8) + \frac{1}{2}(4) + 4 = -\frac{8}{3} + 2 + 4\]

Converting $ 2 $ and $ 4 $ to fractions with a common denominator of $ 3 $:

\[-\frac{8}{3} + \frac{6}{3} + \frac{12}{3} = -\frac{8}{3} + \frac{18}{3} = \frac{10}{3}\]

Thus, the area between the two curves from $ x = 0 $ to $ x = 2 $ is:

\[\text{Area} = \frac{10}{3} \quad \text{or approximately } 3.33\]

This result confirms that even with the absolute value function, we did not need to adjust for negative values since the interval only included positive $ x $ values. The area calculated represents the exact space between the two functions over the specified interval.

concept

Finding Area When Bounds Are Not Given

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Finding Area When Bounds Are Not Given Video Summary

To find the area between two curves when the bounds are not explicitly given, we first need to determine the points where the two functions intersect. This involves setting the functions equal to each other and solving for the variable. For example, consider the functions $ y = 4 - x^2 $ and $ y = 2x + 1 $. To find the intersection points, we set these equations equal:

$ 4 - x^2 = 2x + 1 $

Rearranging this equation leads to:

$ x^2 + 2x - 3 = 0 $

This is a quadratic equation that can be factored as:

$ (x + 3)(x - 1) = 0 $

From this, we find the intersection points at $ x = -3 $ and $ x = 1 $. These points will serve as the bounds for our definite integral.

Next, we set up the integral to calculate the area between the curves. The area $ A $ can be expressed as:

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

We simplify the integrand:

$ A = \int_{-3}^{1} \left( -x^2 - 2x + 3 \right) \, dx $

Now, we find the antiderivative:

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

Applying the Fundamental Theorem of Calculus, we evaluate the antiderivative at the bounds:

First, substituting the upper bound $ x = 1 $:

$ -\frac{1}{3}(1)^3 - (1)^2 + 3(1) = -\frac{1}{3} - 1 + 3 = \frac{5}{3} $

Next, substituting the lower bound $ x = -3 $:

$ -\frac{1}{3}(-3)^3 - (-3)^2 + 3(-3) = -\frac{1}{3}(-27) - 9 - 9 = 9 - 9 - 9 = -9 $

Now, we subtract the lower bound evaluation from the upper bound evaluation:

$ A = \frac{5}{3} - (-9) = \frac{5}{3} + 9 = \frac{5}{3} + \frac{27}{3} = \frac{32}{3} $

Thus, the area between the two curves is $ \frac{32}{3} $ square units. This method of finding intersection points and setting up the integral is essential for calculating areas between curves when bounds are not provided.

Problem

Find the area between $f (x) = x^{2} - 4$ & $g (x) = - x^{2} + 4$ .

10.67

21.33

Problem

Find the area of the shaded region between $f (x) = x^{4} - x^{2}$ & $g of x equals 3 x squared$ .

8.53

17.07

34.13

4.27

Problem

Find the area of the shaded region ONLY that lies between the line $y equals 1$ & $f of x equals sine 2 x$ .

2.43

1.73

$π$

$2 pi$

concept

Finding Area Between Curves that Cross on the Interval

Video duration:

Finding Area Between Curves that Cross on the Interval Video Summary

When finding the area between two curves, it is essential to identify which function is on top and which is on the bottom over the specified interval. In cases where the functions intersect, the area can be calculated by splitting the interval into segments where the top and bottom functions are clearly defined.

Consider the functions $ f(x) = 4 - x^2 $ and $ g(x) = 2x + 1 $ over the interval from $ x = 0 $ to $ x = 2 $. First, we need to determine the points of intersection, which will help us identify where the functions switch positions. In this example, we find that $ f(x) $ is above $ g(x) $ from $ x = 0 $ to $ x = 1 $ and below from $ x = 1 $ to $ x = 2 $.

To calculate the area between the curves, we set up two separate integrals. For the interval from $ x = 0 $ to $ x = 1 $, where $ f(x) $ is on top, the integral is:

\[\int_{0}^{1} \left( f(x) - g(x) \right) \, dx = \int_{0}^{1} \left( (4 - x^2) - (2x + 1) \right) \, dx\]

For the interval from $ x = 1 $ to $ x = 2 $, where $ g(x) $ is now on top, the integral becomes:

\[\int_{1}^{2} \left( g(x) - f(x) \right) \, dx = \int_{1}^{2} \left( (2x + 1) - (4 - x^2) \right) \, dx\]

By evaluating these two integrals and summing their results, we can find the total area between the two curves over the interval from $ x = 0 $ to $ x = 2 $. This method of splitting the area into segments based on the position of the functions is crucial for accurately calculating areas between curves that intersect.

example

Finding Area Between Curves that Cross on the Interval Example 2

Video duration:

Finding Area Between Curves that Cross on the Interval Example 2 Video Summary

To find the area between two functions, $ f(x) $ and $ g(x) $, over a specified interval, we often need to set up multiple integrals, especially when the functions intersect within that interval. In this case, we are calculating the area between $ f(x) $ and $ g(x) $ from $ x = 0 $ to $ x = 2 $. The functions switch positions at $ x = 1 $, necessitating two separate integrals.

The first integral, representing the area from $ x = 0 $ to $ x = 1 $, is set up as:

\[\int_{0}^{1} (f(x) - g(x)) \, dx = \int_{0}^{1} (4 - x^2 - 2x + 1) \, dx\]

This simplifies to:

\[\int_{0}^{1} (-x^2 - 2x + 3) \, dx\]

For the second integral, from $ x = 1 $ to $ x = 2 $, where $ g(x) $ is on top, we have:

\[\int_{1}^{2} (g(x) - f(x)) \, dx = \int_{1}^{2} (2x + 1 - (4 - x^2)) \, dx\]

This simplifies to:

\[\int_{1}^{2} (x^2 + 2x - 3) \, dx\]

Next, we find the antiderivatives for both integrals. For the first integral:

\[\text{Antiderivative} = -\frac{1}{3}x^3 - x^2 + 3x \bigg|_{0}^{1}\]

For the second integral:

\[\text{Antiderivative} = \frac{1}{3}x^3 + x^2 - 3x \bigg|_{1}^{2}\]

Applying the Fundamental Theorem of Calculus, we evaluate the first integral:

\[\left(-\frac{1}{3}(1)^3 - (1)^2 + 3(1)\right) - \left(-\frac{1}{3}(0)^3 - (0)^2 + 3(0)\right) = -\frac{1}{3} - 1 + 3 = \frac{5}{3}\]

For the second integral:

\[\left(\frac{1}{3}(2)^3 + (2)^2 - 3(2)\right) - \left(\frac{1}{3}(1)^3 + (1)^2 - 3(1)\right)\]

Calculating this gives:

\[\left(\frac{8}{3} + 4 - 6\right) - \left(\frac{1}{3} + 1 - 3\right) = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 2\right) = \frac{2}{3} - \left(-\frac{5}{3}\right) = \frac{7}{3}\]

Finally, we combine the results from both integrals to find the total area:

\[\text{Total Area} = \frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4\]

Thus, the area between the two functions $ f(x) $ and $ g(x) $ from $ x = 0 $ to $ x = 2 $ is $ 4 $ square units.

Problem

Find the shaded area between $f (x) = x^{3} + 2 x^{2}$ & $g (x) = x + 2$ .

2.25

3.08

0.42

2.67

Problem

Find the area of the shaded region between $f of x equals sine 2 x$ & $g of x equals 2 sine x$ from $x equals 0$ to $x equals 2 pi$ .

example

Finding Area Between Curves that Cross on the Interval Example 3

Video duration:

Finding Area Between Curves that Cross on the Interval Example 3 Video Summary

To find the areas represented by the shaded regions in the graph, we can set up definite integrals based on the functions involved. Let's break down the three areas: A, B, and C.

For Area A, which is shaded in yellow, we observe that it is bounded by the function g(x) on top and f(x) on the bottom. The interval for this area extends from x = -6 to the intersection point at x = -5. Therefore, the integral representing this area is:

$$ A = \int_{-6}^{-5} (g(x) - f(x)) \, dx $$

Next, we consider Area B, shaded in light red. This area is located beneath the function f(x) and above the x-axis, from x = -4 to x = 0. Since we are calculating the area under a single curve, we do not need to subtract anything. The integral for this area is simply:

$$ B = \int_{-4}^{0} f(x) \, dx $$

Finally, for Area C, shaded in green, we have two segments to consider due to the functions intersecting. The first segment extends from the origin (x = 0) to the intersection point at x = 7, where g(x) is on top and f(x) is on the bottom. The integral for this part is:

$$ C_1 = \int_{0}^{7} (g(x) - f(x)) \, dx $$

For the second segment of Area C, which extends from x = 7 to x = 9, the roles of the functions switch, with f(x) now on top and g(x) on the bottom. The integral for this area is:

$$ C_2 = \int_{7}^{9} (f(x) - g(x)) \, dx $$

To summarize, the three integrals representing the shaded areas are:

$$ A = \int_{-6}^{-5} (g(x) - f(x)) \, dx $$

$$ B = \int_{-4}^{0} f(x) \, dx $$

$$ C = \int_{0}^{7} (g(x) - f(x)) \, dx + \int_{7}^{9} (f(x) - g(x)) \, dx $$

These integrals allow us to calculate the respective areas under the curves defined by the functions.

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9. Graphical Applications of Integrals
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Here’s what students ask on this topic:

How do you find the area between two curves using definite integrals?

To find the area between two curves, you first need to identify the functions and the interval over which you want to calculate the area. The interval is typically determined by the intersection points of the two functions. Once you have the interval, set up a definite integral from the lower bound to the upper bound. The integrand should be the top function minus the bottom function. This means you subtract the function that lies below from the function that lies above within the given interval. If the functions intersect within the interval, you may need to split the integral into two parts, adjusting the top and bottom functions accordingly. This method ensures you accurately calculate the area between the curves.

What do you do if the bounds for finding the area between curves are not given?

If the bounds are not explicitly given, you can determine them by finding the intersection points of the two functions. To do this, set the functions equal to each other and solve for the variable, typically x. These solutions will give you the points of intersection, which serve as the bounds for your definite integral. Once you have these bounds, you can set up the integral of the top function minus the bottom function over the interval defined by these intersection points. This approach allows you to find the area between the curves even when the bounds are not initially provided.

How do you handle finding the area between curves when the top and bottom functions switch?

When the top and bottom functions switch within the interval, you need to split the integral into two separate parts. First, identify the point where the functions intersect and switch positions. Set up one integral from the start of the interval to the intersection point, using the initial top and bottom functions. Then, set up a second integral from the intersection point to the end of the interval, using the new top and bottom functions. Evaluate both integrals separately and add their results to find the total area between the curves. This method ensures you account for the change in positions of the functions.

Can the method for finding the area between curves be used if the functions are below the x-axis?

Yes, the method for finding the area between curves can be used even if the functions are below the x-axis. The key is to focus on the relative positions of the functions to each other, rather than their positions relative to the x-axis. By integrating the top function minus the bottom function over the specified interval, you will obtain the correct area between the curves, regardless of whether they are above or below the x-axis. This approach ensures that the calculated area is always positive and accurately represents the space between the two curves.

What is the role of algebra in finding the area between curves?

Algebra plays a crucial role in finding the area between curves, especially when determining the bounds of integration. When the bounds are not given, algebra is used to find the intersection points of the functions by setting them equal to each other and solving for the variable. This process often involves solving quadratic equations, which may require factoring, using the quadratic formula, or other algebraic techniques. Additionally, algebra is used to simplify the integrand by combining like terms and distributing negatives, ensuring the integral is set up correctly for evaluation. Mastery of algebraic skills is essential for accurately finding the area between curves.

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Area Between Curves: Videos & Practice Problems

Finding Area Between Curves on a Given Interval

Finding Area Between Curves on a Given Interval Video Summary

Calculate the area of the shaded region between the 2 functions from x=0x=0 to x=9x=9

Calculate the area of the shaded region between f(x)f\left(x\right) & g(x)g\left(x\right) contained between x=−4x=-4 & x=−2x=-2.

Sketch the region bounded by f(x)=−(x−2)2+5\textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5} & g(x)=4x\textcolor{orange}{g\left(x\right)=4x} on the interval [0,1]\left\lbrack0,1\right\rbrack. Then set up an integral to represent the region's area and evaluate.

Shade the region bounded by f(x)=sinx\textcolor{blue}{f\left(x\right)=\sin x} & g(x)=cos2x\textcolor{orange}{g\left(x\right)=\cos2x} on the interval [π6,5π6]\left\lbrack\frac{\pi}{6},\frac{5\pi}{6}\right\rbrack. Then set up an integral to represent the region's area.

Finding Area Between Curves on a Given Interval

Finding Area Between Curves on a Given Interval Video Summary

Finding Area When Bounds Are Not Given

Finding Area When Bounds Are Not Given Video Summary

Find the area between f(x)=x2−4f\left(x\right)=x^2-4 & g(x)=−x2+4g\left(x\right)=-x^2+4.

Find the area of the shaded region between f(x)=x4−x2f\left(x\right)=x^4-x^2 & g(x)=3x2g\left(x\right)=3x^2.

Find the area of the shaded region ONLY that lies between the line y=1y=1 & f(x)=sin2xf\left(x\right)=\sin2x.

Finding Area Between Curves that Cross on the Interval

Finding Area Between Curves that Cross on the Interval Video Summary

Finding Area Between Curves that Cross on the Interval Example 2

Finding Area Between Curves that Cross on the Interval Example 2 Video Summary

Find the shaded area between f(x)=x3+2x2\textcolor{orange}{f\left(x\right)=x^3+2x^2} & g(x)=x+2\textcolor{blue}{g\left(x\right)=x+2}.

Find the area of the shaded region between f(x)=sin2x\textcolor{orange}{f\left(x\right)=\sin2x} & g(x)=2sinx\textcolor{blue}{g\left(x\right)=2\sin x} from x=0x=0 to x=2πx=2\pi.

Finding Area Between Curves that Cross on the Interval Example 3

Finding Area Between Curves that Cross on the Interval Example 3 Video Summary

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Go over this topic definitions with flashcards

Here’s what students ask on this topic:

How do you find the area between two curves using definite integrals?

What do you do if the bounds for finding the area between curves are not given?

How do you handle finding the area between curves when the top and bottom functions switch?

Can the method for finding the area between curves be used if the functions are below the x-axis?

What is the role of algebra in finding the area between curves?

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Calculate the area of the shaded region between the 2 functions from $x equals 0$ to $x equals 9$

Calculate the area of the shaded region between $f (x)$ & $g (x)$ contained between $x = - 4$ & $x = - 2$ .

Sketch the region bounded by $f (x) = - {(x - 2)}^{2} + 5$ & $g of x equals 4 x$ on the interval $open bracket 0 comma 1 close bracket$ . Then set up an integral to represent the region's area and evaluate.

Shade the region bounded by $f (x) = \sin x$ & $g of x equals cosine 2 x$ on the interval $[\frac{π}{6}, \frac{5 π}{6}]$ . Then set up an integral to represent the region's area.

Find the area between $f (x) = x^{2} - 4$ & $g (x) = - x^{2} + 4$ .

Find the area of the shaded region between $f (x) = x^{4} - x^{2}$ & $g of x equals 3 x squared$ .

Find the area of the shaded region ONLY that lies between the line $y equals 1$ & $f of x equals sine 2 x$ .

Find the shaded area between $f (x) = x^{3} + 2 x^{2}$ & $g (x) = x + 2$ .

Find the area of the shaded region between $f of x equals sine 2 x$ & $g of x equals 2 sine x$ from $x equals 0$ to $x equals 2 pi$ .