Integration is more than just the reverse of differentiation; it represents the accumulation of quantities. When integrating a function, it is essentially adding up an infinite number of infinitesimally small areas under the curve to find a total area. This concept is fundamental in understanding how to calculate areas under curves. In this section, the focus is on the application of integration to find areas bounded by curves, lines, and between multiple curves.

Area Under a Single Curve

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Area Bounded by the Curve and the x-axis

• Concept: Finding the area between a curve and the x-axis.
• Method: Integrate with respect to dx.
• Procedure: Express y as a function of x, integrate from a to b.

$\int_a^b y \, \mathrm{dx}$

• Example: For the curve $y = x^2$ between $x = 1$ and ( x = 3 ), the area is $\int_1^3 x^2 \, \mathrm{dx}$.

Area Bounded by the Curve and the y-axis

• Concept: Finding the area between a curve and the y-axis.
• Method: Integrate with respect to dy.
• Procedure: Make x the subject, integrate from a to b.

$\int_a^b x \, \mathrm{dy}$

• Example: For the curve $x = y^2$ from $y = 2$ to $y = 4$, the area is $\int_2^4 y^2 \, \mathrm{dy}.$

Example: Specific Area Calculation

Calculate the area under the curve ($\int_2^4 (3x^3 - 4x^2 + 2x + 5) \, \mathrm{dx}.$

Solution:

1. Integrate each term, including the constant as $5x^0$.

2. Substitute the limits to get $\frac{382}{3}$.

Area Between Two Curves

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With Respect to x

• Concept: Calculating the area between two curves along the x-axis.
• Method: Subtract the integral of the lower curve from the upper curve.

$\int_a^b (y_1 - y_2) \, \mathrm{dx}$

• Example: Between curves $y_1 = x^2$ and $y_2 = x$ from $x = 0$ to $x = 1$, calculate $\int_0^1 (x^2 - x) \, \mathrm{dx}$.

With Respect to y

• Method: Similar approach, but with respect to dy.

$\int_a^b (x_1 - x_2) \, \mathrm{dy}$

• Example: Between curves $x_1 = y$ and $x_2 = y^2$ from $y = 0$ to $y = 1$, calculate $\int_0^1 (y - y^2) \, \mathrm{dy}$.

Application Example:

Problem Statement

Given the curve $y = \sqrt{4x + 1} + \frac{9}{\sqrt{4x + 1}}$ and its minimum point M:

1. Find derivatives and integrals.
2. Determine coordinates of M.
3. Calculate the area of the region bounded by the curve, the y-axis, and the line through M parallel to the x-axis.

Solution:

• Differentiation and Integration: Apply the Chain Rule and reverse chain rule.
• Finding M's Coordinates:

Set $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$, solve for x, then find y.

• Area Calculation: Integrate between the curve and the line $y = 6$from $x = 0$ to $x = 2$.