## Quick Math Graphical Integral Calculus

##### Enter expression in this form

Example (common function)

- \( f(x) = 2x + 1 \) enter as 2x +1
- \( f(x) = x^2 - 4x + 3 \) enter as x^2 - 4x + 3
- \( f(x) = \sin(x) \) enter as sin(x)
- \( f(x) = 3 \sin(5x) + 5 \cos(x) \) enter as 3sin(5x) + 5cos(x)

### Graphical Integral Calculus

Graphical integral calculus involves the use of graphs to understand and visualize the concepts of integration. Integration is the process of finding the area under a curve, and it provides a way to accumulate quantities over an interval.

#### Key Concepts

1. Definite Integral: Represents the area under the curve of a function \( f(x) \) from \( x = a \) to \( x = b \).

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

2. Indefinite Integral: Represents the antiderivative of a function, providing a family of functions whose derivative is the integrand.

\[ \int f(x) \, dx = F(x) + C \]

#### Steps in Graphical Integration

1. Plot the Function: Begin by graphing the function \( f(x) \).

2. Identify the Interval: Determine the interval \([a, b]\) over which you need to find the area under the curve.

3. Estimate the Area: The area under the curve from \( x = a \) to \( x = b \) represents the definite integral \(\int_{a}^{b} f(x) \, dx\).

#### Examples

##### Example 1: Area Under a Straight Line

Function:

\[ f(x) = 2x \]

Interval:

\[ x = 0 \text{ to } x = 3 \]

##### Solution:

1. Plot the line \( y = 2x \).

2. Identify the area under the line from \( x = 0 \) to \( x = 3 \).

3. The area forms a right triangle with base 3 and height 6 (since \( f(3) = 6 \)).

4. Calculate the area:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 6 = 9 \]

5. Therefore, \(\int_{0}^{3} 2x \, dx = 9\).

##### Example 2: Area Under a Parabola

Function:

\[ f(x) = x^2 \]

Interval:

\[ x = 1 \text{ to } x = 2 \]

##### Solution:

1. Plot the curve \( y = x^2 \).

2. Identify the area under the curve from \( x = 1 \) to \( x = 2 \).

3. The area under the curve can be approximated using rectangles or trapezoids or calculated exactly using integration.

4. Exact calculation:

\[ \int_{1}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{1}^{2} = \frac{2^3}{3} – \frac{1^3}{3} = \frac{8}{3} – \frac{1}{3} = \frac{7}{3} \]

5. Therefore, the area under \( y = x^2 \) from 1 to 2 is \(\frac{7}{3}\).

##### Example 3: Area Under a Sine Wave

Function:

\[ f(x) = \sin(x) \]

Interval:

\[ x = 0 \text{ to } x = \pi \]

##### Solution:

1. Plot the curve \( y = \sin(x) \).

2. Identify the area under the curve from \( x = 0 \) to \( x = \pi \).

3. The area can be visualized as the region under one-half of the sine wave cycle.

4. Calculate the exact area:

\[ \int_{0}^{\pi} \sin(x) \, dx = \left[ -\cos(x) \right]_{0}^{\pi} = -\cos(\pi) – (-\cos(0)) = 1 – (-1) = 2 \]

5. Therefore, the area under \( y = \sin(x) \) from 0 to \(\pi\) is 2.

#### Practical Applications

– Physics: Graphical integration is used to find displacement from velocity-time graphs, work done from force-displacement graphs, etc.

– Economics: Used to calculate total cost, revenue, and consumer surplus.

– Engineering: Helps in determining quantities like the area under stress-strain curves or the total charge from current-time graphs.

#### Summary

Graphical integral calculus provides a visual and intuitive method to understand and calculate the area under curves. By plotting functions and identifying areas under these curves, we can interpret definite integrals and understand accumulation processes in various applications.