Sketch the region bounded by f ( x ) = − ( x − 2 ) 2 + 5 \textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5} f ( x ) = − ( x − 2 ) 2 + 5 & g ( x ) = 4 x \textcolor{orange}{g\left(x\right)=4x} g ( x ) = 4 x on the interval [ 0 , 1 ] \left\lbrack0,1\right\rbrack [ 0 , 1 ] . 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 ∫ 0 1 − x 2 + 1 d x ; Area = 0.67 0.67 0.67

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

Integral = ; ∫ 0 4 ( − x 2 + 9 ) d x \int_0^4\left(-x^2+9\right)dx ∫ 0 4 ( − x 2 + 9 ) d x Area: 21.33 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 22.67

Shade the region bounded by f ( x ) = l o g 2 x + 4 \textcolor{blue}{f\left(x\right)=log_2x+4} & g ( x ) = e x − 4 \textcolor{orange}{g\left(x\right)=e^{x-4}} on the interval [1,4]. Then set up an integral to represent the region's area.

A = ∫ 1 4 ( l o g 2 x + 4 − e x − 4 ) d x A=\int_1^4\left(log_2x+4-e^{x-4}\right)dx A = ∫ 1 4 ( l o g 2 x + 4 − e x − 4 ) d x

A = ∫ 1 4 ( l o g 2 x + 4 + e x − 4 ) d x A=\int_1^4\left(log_2x+4+e^{x-4}\right)dx A = ∫ 1 4 ( l o g 2 x + 4 + e x − 4 ) d x

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, first identify the intersection points by setting the functions equal to each other. Use definite integrals to calculate the area, integrating the top function minus the bottom function. If the top and bottom functions switch within the interval, split the integral into two parts. The area can be expressed as: $\int_{a}^{b} (f (x) - g (x)) d x$ .

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 to visualize the area of interest. For example, if we want to find the area between $ f(x) $ and $ g(x) $ from $ x = 0 $ to $ x = 1 $, we first identify which function is on top and which is on the bottom within that interval.

The area between the curves can be determined by calculating the area under the top function and subtracting the area under the bottom function. This can be expressed mathematically as:

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

By combining these integrals, we can simplify the expression to:

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

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

$$ A = \int_{0}^{1} ((4 - x^2) - (2x + 1)) \, dx $$

After simplifying the integrand, we rewrite it in standard form:

$$ A = \int_{0}^{1} (-x^2 - 2x + 3) \, dx $$

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

$$ A = \left[-\frac{1}{3}x^3 - x^2 + 3x\right]_{0}^{1} $$

Evaluating this expression at the bounds $ 0 $ and $ 1 $ gives:

$$ A = \left(-\frac{1}{3}(1)^3 - (1)^2 + 3(1)\right) - \left(-\frac{1}{3}(0)^3 - (0)^2 + 3(0)\right) $$

After performing the calculations, we find:

$$ A = -\frac{1}{3} - 1 + 3 = \frac{5}{3} $$

This result, $ \frac{5}{3} $, represents the area between the two curves over the specified interval. It is important to note that this method remains valid even if the functions lie 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 $x=0$ to $x=9$

4.5

445

Problem

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

8.17

4.17

-12.17

0.50

Problem

Sketch the region bounded by $\textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5}$ & $\textcolor{orange}{g\left(x\right)=4x}$ on the interval $\left\lbrack0,1\right\rbrack$ . 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 = $\int_0^1\left(-x^2+9\right)dx$ ; 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 = $\int_0^1-x^2+1dx$ ; 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 = ; $\int_0^4\left(-x^2+9\right)dx$ 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 $f (x) = l o g_{2} x + 4$ & $g (x) = e^{x - 4}$ on the interval [1,4]. Then set up an integral to represent the region's area.

Graph showing the area between the curves f(x) and g(x) shaded in yellow on the interval [1,4].

$A=\int_1^4\left(log_2x+4-e^{x-4}\right)dx$

Graph showing the area between the curves f(x) and g(x) shaded in yellow, bounded on the interval [1,4].

$A=\int_1^4\left(log_2x+4-e^{x-4}\right)dx$

Graph showing the area between the curves f(x) and g(x) shaded in yellow, with labeled axes and functions.

$A=\int_1^4\left(log_2x+4+e^{x-4}\right)dx$

Graph showing the area between the curves f(x) and g(x) shaded in yellow on the interval [1,4].

$A=\int_1^4\left(log_2x+4+e^{x-4}\right)dx$

example

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 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.

The area $ A $ can be expressed as:

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

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

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

Rearranging the integrand gives us:

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

Next, we find the antiderivative of the function:

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

Now, we apply the Fundamental Theorem of Calculus by evaluating the antiderivative at the bounds $ 0 $ and $ 2 $:

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

Calculating the upper bound:

\[= -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} = \frac{10}{3}\]

Since the lower bound evaluates to zero, we have:

\[A = \frac{10}{3}\]

Thus, the area between the two functions from $ x = 0 $ to $ x = 2 $ is $ \frac{10}{3} $, which is approximately $ 3.33 $ when expressed as a decimal. This calculation demonstrates that even with the absolute value function, we can simplify our work by recognizing the positivity of the interval.

concept

Finding Area When Bounds Are Not Given

Video duration:

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 factorization, 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} (-x^2 - 2x + 3) \, dx $

Now, we find the antiderivative of the integrand:

$ 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 process illustrates how to find the area between two curves by determining their intersection points and setting up a definite integral accordingly.

Problem

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

10.67

21.33

Problem

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

8.53

17.07

34.13

4.27

Problem

Find the area of the shaded region shown in the first quadrant between $f (x) = \frac{1}{x}$ & $g (x) = - \frac{1}{4} x + \frac{5}{4} .$

4.239

1.386

0.489

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 top and bottom functions may change within the interval. To accurately calculate the area in such scenarios, we can split the area into two separate integrals, each corresponding to the segments of the interval 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 0 to 2. By graphing these functions, we can observe their behavior:

From $ x = 0 $ to $ x = 1 $, $ f(x) $ is above $ g(x) $.
From $ x = 1 $ to $ x = 2 $, $ g(x) $ takes the position above $ f(x) $.

To find the total area between the curves, we set up two integrals:

1. For the interval from 0 to 1, where $ f(x) $ is the top function:

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

2. For the interval from 1 to 2, where $ g(x) $ is now the top function:

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

By evaluating these two integrals and summing their results, we can determine the total area between the two curves over the interval from 0 to 2. This method ensures that we account for the changing positions of the functions accurately, allowing for a precise calculation of the area.

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 first identify where the functions intersect. In this case, we are calculating the area between these functions from $ x = 0 $ to $ x = 2 $. Notably, the functions switch positions within this interval, necessitating the use of two separate integrals to accurately capture the area.

The first integral, from $ 0 $ to $ 1 $, is set up as follows:

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

This simplifies to:

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

For the second integral, from $ 1 $ to $ 2 $, we have:

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

This simplifies to:

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

Next, we find the antiderivatives for both integrals. The antiderivative for the first integral is:

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

And for the second integral, it is:

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

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

Plugging in the upper bound $ x = 1 $:

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

For the lower bound $ x = 0 $:

\[0\]

Thus, the result of the first integral is:

\[\frac{5}{3} - 0 = \frac{5}{3}\]

Now, evaluating the second integral:

For the upper bound $ x = 2 $:

\[\frac{1}{3}(2)^3 + (2)^2 - 3(2) = \frac{8}{3} + 4 - 6 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}\]

For the lower bound $ x = 1 $:

\[\frac{1}{3}(1)^3 + (1)^2 - 3(1) = \frac{1}{3} + 1 - 3 = \frac{1}{3} - \frac{9}{3} = -\frac{8}{3}\]

Thus, the result of the second integral is:

\[\frac{2}{3} - (-\frac{8}{3}) = \frac{2}{3} + \frac{8}{3} = \frac{10}{3}\]

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

\[\text{Total Area} = \frac{5}{3} + \frac{10}{3} = \frac{15}{3} = 5\]

Therefore, the area between the two functions $ f(x) $ and $ g(x) $ from $ x = 0 $ to $ x = 2 $ is $ 5 $.

Problem

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

2.25

3.08

0.42

2.67

Problem

Find the area of the shaded region between $f (x) = \frac{1}{x}$ & $g (x) = x$ from $x equals 0.5$ to $x equals 4$ .

-5.796

6.432

0.329

0.557

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. The integral representing this area is:

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

This integral calculates the area between the two curves over the specified interval.

Next, for Area B, shaded in light red, this area is located beneath the function f(x) and above the x-axis. Since the x-axis is represented by y = 0, we do not need to subtract anything from f(x). The integral for this area, from x = -4 to the origin at x = 0, is simply:

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

This integral gives us the area under the curve of f(x) within the specified limits.

Finally, for Area C, shaded in green, we have two segments to consider due to the intersection of the functions. The first segment extends from the origin 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, 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 find the total area C, we would sum these two integrals:

$$ C = C_1 + C_2 = \int_{0}^{7} (g(x) - f(x)) \, dx + \int_{7}^{9} (f(x) - g(x)) \, dx $$

In summary, the three integrals representing the shaded areas are:

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

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

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

Do you want more practice?

We have more practice problems on Area Between Curves

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, first identify the intersection points of the curves to establish the bounds for integration. Once the bounds are determined, set up the integral as the top function minus the bottom function over the interval. The integral is expressed as:

\int a b (f (x) - g (x)) d x

If the functions switch positions within the interval, split the integral into two parts, each with its respective top and bottom functions. Evaluate the integrals to find the area accurately.

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

If the bounds for integration are not given, you can find them by determining the intersection points of the two curves. Set the functions equal to each other and solve for $x$ to find these points. These intersection points will serve as the bounds for your integral. Once you have the bounds, set up the integral as the top function minus the bottom function over the interval. This method ensures you accurately calculate the area between the curves.

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

When the top and bottom functions switch within the interval, split the integral into two separate integrals at the intersection point where the switch occurs. For each segment, set up the integral as the top function minus the bottom function. Evaluate each integral separately and then sum the results to find the total area. This approach ensures that you account for the change in position of the functions and accurately calculate the area between them.

Can you find the area between curves if the functions are below the x-axis?

Yes, you can find the area between curves even if the functions are below the x-axis. The process remains the same: identify the intersection points to establish the bounds, and set up the integral as the top function minus the bottom function. The position of the functions relative to the x-axis does not affect the calculation of the area between them, as the integral accounts for the vertical distance between the curves.

What is the significance of the intersection points when finding the area between curves?

The intersection points are crucial when finding the area between curves because they determine the bounds for the integral. These points indicate where the curves meet and help define the interval over which you will integrate. By setting the functions equal to each other and solving for $x$ , you can find these intersection points and use them to accurately calculate the area between the curves.

Your Business Calculus tutors

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=0x=0 to x=9x=9x=9

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

Shade the region bounded by f(x)=log2x+4\textcolor{blue}{f\left(x\right)=log_2x+4} & g(x)=ex−4\textcolor{orange}{g\left(x\right)=e^{x-4}} on the interval [1,4]. 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^2-4f(x)=x2−4 & g(x)=−x2+4g\left(x\right)=-x^2+4g(x)=−x2+4.

Find the area of the shaded region between f(x)=x4−x2f\left(x\right)=x^4-x^2f(x)=x4−x2 & g(x)=3x2g\left(x\right)=3x^2g(x)=3x2.

Find the area of the shaded region shown in the first quadrant between f(x)=1xf\left(x\right)=\frac{1}{x} & g(x)=−14x+54.g\left(x\right)=-\frac14x+\frac54.

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}f(x)=x3+2x2 & g(x)=x+2\textcolor{blue}{g\left(x\right)=x+2}g(x)=x+2.

Find the area of the shaded region between f(x)=1xf\left(x\right)=\frac{1}{x} & g(x)=xg\left(x\right)=x from x=0.5x=0.5 to x=4x=4.

Finding Area Between Curves that Cross on the Interval Example 3

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

Do you want more practice?

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 integration are not given when finding the area between curves?

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

Can you find the area between curves if the functions are below the x-axis?

What is the significance of the intersection points when finding the area between curves?

Your Business Calculus tutors

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

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

Shade the region bounded by $f (x) = l o g_{2} x + 4$ & $g (x) = e^{x - 4}$ on the interval [1,4]. Then set up an integral to represent the region's area.

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

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

Find the area of the shaded region shown in the first quadrant between $f (x) = \frac{1}{x}$ & $g (x) = - \frac{1}{4} x + \frac{5}{4} .$

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

Find the area of the shaded region between $f (x) = \frac{1}{x}$ & $g (x) = x$ from $x equals 0.5$ to $x equals 4$ .