Maharashtra Board Class 12 Maths Part 2 Chapter 5 Application Of Definite Integration PDF Download

Application Of Definite Integration

Let Us Study

Area under the curve

Area bounded by the curve, axis and given lines

Area between two curves

Let Us Recall

In previous chapter we have studied definition of definite integral as limit of a sum. Geometrically \(\int_a^b f(x) \cdot dx\) gives the area A under the curve \(y = f(x)\) with \(f(x) \geq 0\) and bounded by the X-axis and the lines \(x = a\), \(x = b\); and is given by

\[\int_a^b f(x) \, dx = \phi(b) - \phi(a)\]

where \(\int f(x) \, dx = \phi(x)\)

The curve \(y = f(x)\) is continuous in \([a, b]\) and \(f(x) \geq 0\) in \([a, b]\).

This is also known as fundamental theorem of integral calculus. We shall find the area under the curve by using definite integral.

5.1 Area Under The Curve

For evaluation of area bounded by certain curves, we need to know the nature of the curves and their graphs. We should also be able to draw sketch of the curves.

5.1.1 Area Under A Curve

The area shaded in figure is bounded by the curve \(y = f(x)\), X-axis and the lines \(x = a\), \(x = b\) and is given by the definite integral \(\int_{x=a}^{x=b} (y) \cdot dx\)

A = area of the shaded region.

\[A = \int_a^b f(x) \cdot dx\]

Teacher's Note

This chapter teaches us how to find areas of shapes using calculus. For example, if you want to find the area under a bridge or under a curve in a garden, you can use this method.

Exam Trick

Remember: Integration helps you find area. Just like you count small pieces to find total cost, integration adds up tiny strips to find total area.

Points To Remember

Definite integral gives the area under a curve between two lines.
The curve must be continuous and above the X-axis.
The area is always positive.
We use the formula A = φ(b) - φ(a) where φ is the antiderivative.

The area A, bounded by the curve \(x = g(y)\), Y axis and the lines \(y = c\) and \(y = d\) is given by

\[A = \int_{y=c}^{y=d} x \cdot dy = \int_{y=c}^{y=d} g(y) \cdot dx\]

Solved Example

Ex. 1 : Find the area bounded by the curve \(y = x\), the Y axis the X axis and \(x = 3\).

Solution : The required area \(A = \int_0^3 y \cdot dx\)

\[A = \int_0^3 x \cdot dx\]

\[= \left[ \frac{x^3}{3} \right]_0^3\]

\[A = 9 - 0 = 9 \text{ sq.units}\]

Teacher's Note

This example shows a simple area calculation. Imagine a triangle in a shop where you need to find how much space it takes on the floor.

Exam Trick

Always find the antiderivative first, then put the upper and lower limits. The answer will be in square units.

Points To Remember

First substitute the upper limit in the antiderivative.
Then subtract the result when you substitute the lower limit.
Always write the answer with square units (sq.units).
Check if the curve is above or below the X-axis.

5.1.2 Area Between Two Curves

Let \(y = f(x)\) and \(y = g(x)\) be the equations of the two curves as shown in figure.

Let A be the area bounded by the curves \(y = f(x)\) and \(y = g(x)\)

\[A = | A_1 - A_2 |\]

where

\(A_1\) = Area bounded by the curve \(y = f(x)\), X-axis and \(x = a\), \(x = b\).

\(A_2\) = Area bounded by the curve \(y = g(x)\), X-axis and \(x = a\), \(x = b\).

The point of intersection of the curves \(y = f(x)\) and \(y = g(x)\) can be obtained by solving their equations simultaneously.

Therefore the required area \[A = \left| \int_a^b f(x) \, dx - \int_a^b g(x) \, dx \right|\]

Teacher's Note

When two curves overlap, we find the area between them by subtracting. Like finding how much more milk one bottle has than another bottle.

Exam Trick

First find where the two curves meet (intersection points). These become your new limits for integration.

Points To Remember

Always subtract the lower curve from the upper curve.
Find intersection points by solving both equations together.
Use absolute value to keep area positive.
The area is between the two curves, not including extra spaces.

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