# Nakafa Framework: LLM

URL: https://nakafa.com/en/subject/high-school/12/mathematics/integral/area-of-a-flat-surface
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/12/mathematics/integral/area-of-a-flat-surface/en.mdx

Output docs content for large language models.

---

export const metadata = {
  title: "Area of a Flat Surface",
  description: "Calculate flat surface areas using definite integrals with step-by-step solutions. Master quadratic and irrational functions through practical examples.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/26/2025",
  subject: "Integrals",
};

import { LineEquation } from "@repo/design-system/components/contents/line-equation";
import { getColor } from "@repo/design-system/lib/color";

## 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 <InlineMath math="f(x)" /> and want to find the area of the region under the curve from <InlineMath math="x = a" /> to <InlineMath math="x = b" />. We can divide the interval <InlineMath math="[a, b]" /> into <InlineMath math="n" /> small parts with width <InlineMath math="\Delta x" />.

<div className="flex flex-col gap-4">
<BlockMath math="A = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \cdot \Delta x" />
<BlockMath math="A = \int_{a}^{b} f(x) \, dx" />
</div>

## 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 <InlineMath math="x" /> 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 <InlineMath math="x" />-axis, then the area of the region is <InlineMath math="\int_{a}^{b} f(x) \, dx" />.

### Evaluate the Integral

After determining the limits and function, we can evaluate the integral using the fundamental theorem of calculus:

<BlockMath math="\int_{a}^{b} f(x) \, dx = F(b) - F(a)" />

where <InlineMath math="F(x)" /> is the antiderivative of <InlineMath math="f(x)" />.

## 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 <InlineMath math="f(x) = x^2 - 4x" /> and the <InlineMath math="x" />-axis between <InlineMath math="x = 1" /> and <InlineMath math="x = 3" />.

<LineEquation
  title={<>Graph of Function <InlineMath math="f(x) = x^2 - 4x" /></>}
  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 <InlineMath math="f(x) = x^2 - 4x" /> turns out to have negative values in the interval <InlineMath math="[1, 3]" />. We can easily check: when <InlineMath math="x = 1" />, we get <InlineMath math="f(1) = 1 - 4 = -3" />. Similarly, when <InlineMath math="x = 3" />, we get <InlineMath math="f(3) = 9 - 12 = -3" />.

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:

<div className="flex flex-col gap-4">
<BlockMath math="A = \int_{1}^{3} |x^2 - 4x| \, dx" />
<BlockMath math="A = \int_{1}^{3} -(x^2 - 4x) \, dx" />
<BlockMath math="A = \int_{1}^{3} (-x^2 + 4x) \, dx" />
</div>

Let's solve it step by step:

<div className="flex flex-col gap-4">
<BlockMath math="A = \left[-\frac{x^3}{3} + 2x^2\right]_{1}^{3}" />
<BlockMath math="A = \left(-\frac{27}{3} + 18\right) - \left(-\frac{1}{3} + 2\right)" />
<BlockMath math="A = 9 - \frac{5}{3} = \frac{22}{3}" />
</div>

Therefore, the area of that region is <InlineMath math="\frac{22}{3}" /> 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 <InlineMath math="f(x) = x\sqrt{x^2 + 5}" /> from <InlineMath math="x = 0" /> to <InlineMath math="x = 2" />.

<LineEquation
  title={<>Graph of Function <InlineMath math="f(x) = x\sqrt{x^2 + 5}" /></>}
  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 <InlineMath math="x\sqrt{x^2 + 5}" /> which is quite complex if we solve it directly.

Let's perform substitution with <InlineMath math="u = x^2 + 5" />. From here, we get the differential <InlineMath math="du = 2x \, dx" />, which means <InlineMath math="x \, dx = \frac{1}{2} du" />.

Don't forget to change the integration limits too! When <InlineMath math="x = 0" />, we get <InlineMath math="u = 5" />. When <InlineMath math="x = 2" />, we get <InlineMath math="u = 9" />.

Now our integral becomes:

<div className="flex flex-col gap-4">
<BlockMath math="A = \int_{0}^{2} x\sqrt{x^2 + 5} \, dx" />
<BlockMath math="A = \int_{5}^{9} \frac{1}{2}\sqrt{u} \, du" />
<BlockMath math="A = \frac{1}{2} \int_{5}^{9} u^{1/2} \, du" />
<BlockMath math="A = \frac{1}{2} \left[\frac{2}{3}u^{3/2}\right]_{5}^{9}" />
<BlockMath math="A = \frac{1}{3}\left[u^{3/2}\right]_{5}^{9}" />
<BlockMath math="A = \frac{1}{3}\left(27 - 5\sqrt{5}\right)" />
</div>

Note that <InlineMath math="9^{3/2} = (3^2)^{3/2} = 3^3 = 27" /> and <InlineMath math="5^{3/2} = 5 \cdot \sqrt{5}" />.

> 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 <InlineMath math="y = x^2 + 1" />, the <InlineMath math="x" />-axis, and the lines <InlineMath math="x = 0" /> and <InlineMath math="x = 2" />!

2. Determine the area of the region under the curve <InlineMath math="y = \frac{1}{x^2 + 1}" /> from <InlineMath math="x = 0" /> to <InlineMath math="x = 1" />!

3. Calculate the area of the region bounded by the curve <InlineMath math="y = 2x - x^2" /> and the <InlineMath math="x" />-axis!

### Answer Key

1. **First problem with function <InlineMath math="y = x^2 + 1" />**

   Since this function is always positive, we can directly set up the integral:
   
   <BlockMath math="A = \int_{0}^{2} (x^2 + 1) \, dx" />

   After we integrate and evaluate, we obtain:
   
   <div className="flex flex-col gap-4">
   <BlockMath math="A = \left[\frac{x^3}{3} + x\right]_{0}^{2}" />
   <BlockMath math="A = \left(\frac{8}{3} + 2\right) - 0 = \frac{14}{3}" />
   </div>

   Therefore, the area of that region is <InlineMath math="\frac{14}{3}" /> square units.

2. **Second problem with rational function**

   For this integral, we need to remember that the antiderivative of <InlineMath math="\frac{1}{x^2 + 1}" /> is <InlineMath math="\arctan(x)" />.

   <div className="flex flex-col gap-4">
   <BlockMath math="A = \int_{0}^{1} \frac{1}{x^2 + 1} \, dx" />
   <BlockMath math="A = [\arctan(x)]_{0}^{1}" />
   <BlockMath math="A = \arctan(1) - \arctan(0) = \frac{\pi}{4}" />
   </div>

   The area of that region is <InlineMath math="\frac{\pi}{4}" /> square units.

3. **Third problem with parabola**

   First, let's find where the curve intersects the <InlineMath math="x" />-axis:
   
   <BlockMath math="2x - x^2 = 0 \Rightarrow x(2 - x) = 0" />

   So the intersection points are at <InlineMath math="x = 0" /> and <InlineMath math="x = 2" />. Since this function is positive between these two points, we can directly integrate:
   
   <div className="flex flex-col gap-4">
   <BlockMath math="A = \int_{0}^{2} (2x - x^2) \, dx" />
   <BlockMath math="A = \left[x^2 - \frac{x^3}{3}\right]_{0}^{2}" />
   <BlockMath math="A = 4 - \frac{8}{3} = \frac{4}{3}" />
   </div>

   The area of that region is <InlineMath math="\frac{4}{3}" /> square units.