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URL: https://nakafa.com/en/subjects/mathematics/integral/area-of-a-flat-surface
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Calculate flat surface areas using definite integrals with worked solutions. Learn quadratic and irrational functions through practical examples.
---
## Basic Concepts of Flat Surface Area
In everyday life, we often need to calculate the area of various flat shapes. For shapes with simple forms like squares or triangles, we can use familiar formulas. But what if we want to calculate the area of a region bounded by irregular curves?
**Definite integrals** provide an elegant solution to this problem. The basic concept of definite integrals stems from Riemann's approach, where we divide a region into small rectangles and then sum their areas.
Imagine we have a function $$f(x)$$ and want to find the area of the region under the curve from $$x = a$$ to $$x = b$$. We can divide the interval $$[a, b]$$ into $$n$$ small parts with width $$\Delta x$$.
Visible text: Imagine we have a function and want to find the area of the region under the curve from to . We can divide the interval into small parts with width .
Component: MathContainer
Children:
```math
A = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \cdot \Delta x
```
```math
A = \int_{a}^{b} f(x) \, dx
```
## Determining Area Using Definite Integrals
To calculate the area of a flat surface using definite integrals, we need to understand several systematic steps:
### Identify Integration Limits
The first step is to determine the lower and upper limits of integration. These limits indicate the range of $$x$$ values that bound the region whose area we want to calculate.
Visible text: The first step is to determine the lower and upper limits of integration. These limits indicate the range of values that bound the region whose area we want to calculate.
### Determine the Integrand Function
The function to be integrated is the function that bounds the region. If the region is above the $$x$$-axis, then the area of the region is $$\int_{a}^{b} f(x) \, dx$$.
Visible text: The function to be integrated is the function that bounds the region. If the region is above the -axis, then the area of the region is .
### Evaluate the Integral
After determining the limits and function, we can evaluate the integral using the fundamental theorem of calculus:
```math
\int_{a}^{b} f(x) \, dx = F(b) - F(a)
```
where $$F(x)$$ is the antiderivative of $$f(x)$$.
Visible text: where is the antiderivative of .
## Application to Quadratic Functions
Let's apply this concept to a concrete example. Suppose we want to calculate the area of the region bounded by the curve $$f(x) = x^2 - 4x$$ and the $$x$$-axis between $$x = 1$$ and $$x = 3$$.
Visible text: Let's apply this concept to a concrete example. Suppose we want to calculate the area of the region bounded by the curve and the -axis between and .
Component: LineEquation
Props:
- title: Graph of Function $$f(x) = x^2 - 4x$$
Visible text: Graph of Function
- description: Visualization of the region whose area will be calculated with the help of boundary lines and shaded area.
- showZAxis: false
- cameraPosition: [0, 0, 12]
- data: [
{
points: Array.from({ length: 31 }, (_, i) => {
const x = 0 + (i * 4) / 30;
const y = x * x - 4 * x;
return { x, y, z: 0 };
}),
color: getColor("PURPLE"),
smooth: true,
showPoints: false,
labels: [
{ text: "f(x) = x² - 4x", at: 25, offset: [0, 3, 0] }
]
},
{
points: [
{ x: 1, y: 0, z: 0 },
{ x: 1, y: -3, z: 0 }
],
color: getColor("ORANGE"),
showPoints: false,
labels: [
{ text: "x = 1", at: 0, offset: [0.3, -1.5, 0] }
]
},
{
points: [
{ x: 3, y: 0, z: 0 },
{ x: 3, y: -3, z: 0 }
],
color: getColor("ORANGE"),
showPoints: false,
labels: [
{ text: "x = 3", at: 0, offset: [0.3, -1.5, 0] }
]
},
{
points: [
{ x: 1, y: -3, z: 0 },
{ x: 3, y: -3, z: 0 }
],
color: getColor("AMBER"),
showPoints: false,
lineWidth: 1,
labels: [
{ text: "Area Region", at: 0, offset: [0, -1.5, 0] }
]
}
]
Now, try to observe the graph above. The function $$f(x) = x^2 - 4x$$ turns out to have negative values in the interval $$[1, 3]$$. We can easily check: when $$x = 1$$, we get $$f(1) = 1 - 4 = -3$$. Similarly, when $$x = 3$$, we get $$f(3) = 9 - 12 = -3$$.
Visible text: Now, try to observe the graph above. The function turns out to have negative values in the interval . We can easily check: when , we get . Similarly, when , we get .
Well, here's where the uniqueness lies! Since we're looking for **area** which is always positive, we need to use the absolute value of that function. So our integral becomes:
Component: MathContainer
Children:
```math
A = \int_{1}^{3} |x^2 - 4x| \, dx
```
```math
A = \int_{1}^{3} -(x^2 - 4x) \, dx
```
```math
A = \int_{1}^{3} (-x^2 + 4x) \, dx
```
Let's solve it step by step:
Component: MathContainer
Children:
```math
A = \left[-\frac{x^3}{3} + 2x^2\right]_{1}^{3}
```
```math
A = \left(-\frac{27}{3} + 18\right) - \left(-\frac{1}{3} + 2\right)
```
```math
A = 9 - \frac{5}{3} = \frac{22}{3}
```
Therefore, the area of that region is $$\frac{22}{3}$$ square units.
Visible text: Therefore, the area of that region is square units.
## Application to Irrational Functions
Now let's try a slightly more challenging example with an irrational function. We will calculate the area of the region under the curve $$f(x) = x\sqrt{x^2 + 5}$$ from $$x = 0$$ to $$x = 2$$.
Visible text: Now let's try a slightly more challenging example with an irrational function. We will calculate the area of the region under the curve from to .
Component: LineEquation
Props:
- title: Graph of Function $$f(x) = x\sqrt{x^2 + 5}$$
Visible text: Graph of Function
- description: Region under the curve whose area will be calculated with interval helper lines.
- showZAxis: false
- data: [
{
points: Array.from({ length: 31 }, (_, i) => {
const x = 0 + (i * 2) / 30;
const y = x * Math.sqrt(x * x + 5);
return { x, y, z: 0 };
}),
color: getColor("EMERALD"),
smooth: true,
showPoints: false,
labels: [
{ text: "f(x) = x√(x² + 5)", at: 25, offset: [-1, -2, 0] }
]
},
{
points: [
{ x: 0, y: 0, z: 0 },
{ x: 0, y: 0, z: 0 }
],
color: getColor("VIOLET"),
showPoints: true,
labels: [
{ text: "x = 0", at: 0, offset: [-0.5, -0.5, 0] }
]
},
{
points: [
{ x: 2, y: 0, z: 0 },
{ x: 2, y: 6, z: 0 }
],
color: getColor("VIOLET"),
showPoints: false,
labels: [
{ text: "x = 2", at: 0, offset: [0.3, -0.5, 0] }
]
},
{
points: [
{ x: 2, y: 6, z: 0 },
{ x: 2, y: 6, z: 0 }
],
color: getColor("VIOLET"),
showPoints: true,
labels: [
{ text: "f(2) = 6", at: 0, offset: [1.5, -1, 0] }
]
}
]
For this integral, we need to use the **substitution technique**. Why? Because there's the form $$x\sqrt{x^2 + 5}$$ which is quite complex if we solve it directly.
Visible text: For this integral, we need to use the **substitution technique**. Why? Because there's the form which is quite complex if we solve it directly.
Let's perform substitution with $$u = x^2 + 5$$. From here, we get the differential $$du = 2x \, dx$$, which means $$x \, dx = \frac{1}{2} du$$.
Visible text: Let's perform substitution with . From here, we get the differential , which means .
Don't forget to change the integration limits too! When $$x = 0$$, we get $$u = 5$$. When $$x = 2$$, we get $$u = 9$$.
Visible text: Don't forget to change the integration limits too! When , we get . When , we get .
Now our integral becomes:
Component: MathContainer
Children:
```math
A = \int_{0}^{2} x\sqrt{x^2 + 5} \, dx
```
```math
A = \int_{5}^{9} \frac{1}{2}\sqrt{u} \, du
```
```math
A = \frac{1}{2} \int_{5}^{9} u^{1/2} \, du
```
```math
A = \frac{1}{2} \left[\frac{2}{3}u^{3/2}\right]_{5}^{9}
```
```math
A = \frac{1}{3}\left[u^{3/2}\right]_{5}^{9}
```
```math
A = \frac{1}{3}\left(27 - 5\sqrt{5}\right)
```
Note that $$9^{3/2} = (3^2)^{3/2} = 3^3 = 27$$ and $$5^{3/2} = 5 \cdot \sqrt{5}$$.
Visible text: Note that and .
> When using substitution in definite integrals, don't forget to change the integration limits according to the new substitution variable.
## Exercises
1. Calculate the area of the region bounded by the curve $$y = x^2 + 1$$, the $$x$$-axis, and the lines $$x = 0$$ and $$x = 2$$!
2. Determine the area of the region under the curve $$y = \frac{1}{x^2 + 1}$$ from $$x = 0$$ to $$x = 1$$!
3. Calculate the area of the region bounded by the curve $$y = 2x - x^2$$ and the $$x$$-axis!
Visible text: 1. Calculate the area of the region bounded by the curve , the -axis, and the lines and !
2. Determine the area of the region under the curve from to !
3. Calculate the area of the region bounded by the curve and the -axis!
### Answer Key
1. **First problem with function $$y = x^2 + 1$$**
Since this function is always positive, we can directly set up the integral:
```math
A = \int_{0}^{2} (x^2 + 1) \, dx
```
After we integrate and evaluate, we obtain:
```math
A = \left[\frac{x^3}{3} + x\right]_{0}^{2}
```
```math
A = \left(\frac{8}{3} + 2\right) - 0 = \frac{14}{3}
```
Therefore, the area of that region is $$\frac{14}{3}$$ square units.
2. **Second problem with rational function**
For this integral, we need to remember that the antiderivative of $$\frac{1}{x^2 + 1}$$ is $$\arctan(x)$$.
```math
A = \int_{0}^{1} \frac{1}{x^2 + 1} \, dx
```
```math
A = [\arctan(x)]_{0}^{1}
```
```math
A = \arctan(1) - \arctan(0) = \frac{\pi}{4}
```
The area of that region is $$\frac{\pi}{4}$$ square units.
3. **Third problem with parabola**
First, let's find where the curve intersects the $$x$$-axis:
```math
2x - x^2 = 0 \Rightarrow x(2 - x) = 0
```
So the intersection points are at $$x = 0$$ and $$x = 2$$. Since this function is positive between these two points, we can directly integrate:
```math
A = \int_{0}^{2} (2x - x^2) \, dx
```
```math
A = \left[x^2 - \frac{x^3}{3}\right]_{0}^{2}
```
```math
A = 4 - \frac{8}{3} = \frac{4}{3}
```
The area of that region is $$\frac{4}{3}$$ square units.
Visible text: 1. **First problem with function **
Since this function is always positive, we can directly set up the integral:
After we integrate and evaluate, we obtain:
Therefore, the area of that region is square units.
2. **Second problem with rational function**
For this integral, we need to remember that the antiderivative of is .
The area of that region is square units.
3. **Third problem with parabola**
First, let's find where the curve intersects the -axis:
So the intersection points are at and . Since this function is positive between these two points, we can directly integrate:
The area of that region is square units.