CLASS XII MATHEMATICS CHAPTER 8

Application of Integrals

Complete Class XII Mathematics chapter notes covering area under curves, area bounded by curves and coordinate axes, area between two curves, vertical and horizontal strips, symmetry and important applications of definite integrals.

CBSE Pattern Definite Integrals Area Under Curves Area Between Curves Graphical Approach HOT Questions

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Chapter Overview

Introduction

Application of Integrals deals with the geometrical interpretation of definite integrals. The principal application in this chapter is the calculation of the area of regions bounded by curves and coordinate axes.

Instead of treating integration only as an algebraic operation, we interpret it geometrically as the accumulation of infinitely many small areas.

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Integration

Accumulation of infinitely small quantities.

▰

Area

Definite integrals can represent geometrical area.

📈

Curves

Areas may be bounded by one or more curves.

Fundamental Application

Area Under a Curve

Let \(y=f(x)\) be a continuous function which remains above the \(x\)-axis between \(x=a\) and \(x=b\). Then the area bounded by the curve, \(x=a\), \(x=b\) and the \(x\)-axis is:

\[ \boxed{ A=\int_a^b f(x)\,dx } \]

Geometrical Interpretation

Consider a very small vertical strip of width \(dx\). Its height is approximately \(f(x)\).

\[ dA=f(x)\,dx \]

Adding all such infinitely thin strips gives the complete area:

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

Key Idea

Integration converts infinitely many infinitesimal strips into the total area.

Area and Sign

Area with the x-axis

Curve Above x-axis

\[ \boxed{ A=\int_a^b f(x)\,dx } \]

Curve Below x-axis

If \(f(x)<0\), the definite integral is negative, but geometrical area is positive.

\[ \boxed{ A=-\int_a^b f(x)\,dx } \]

Curve Crossing the x-axis

If the curve changes sides of the \(x\)-axis, divide the interval at its points of intersection with the axis.

\[ \boxed{ A=\int_a^b |f(x)|\,dx } \]

⚠ Important

A definite integral represents signed area. Geometrical area cannot be negative.

Horizontal Strip

Area with the y-axis

When a region is more conveniently described in terms of \(y\), use horizontal strips.

\[ dA=x\,dy \]

If the curve is \(x=g(y)\), then:

\[ \boxed{ A=\int_c^d g(y)\,dy } \]

Strip

Horizontal

\[ dy \]

Width

Right − Left

\[ x_R-x_L \]

Area

\[ dA=(x_R-x_L)dy \]

Most Important Concept

Area Between Two Curves

Let two curves be:

\[ y=f(x) \] and \[ y=g(x) \]

If \(f(x)\) is above \(g(x)\), then the vertical distance between the curves is:

\[ f(x)-g(x) \]

\[ \boxed{ A=\int_a^b [f(x)-g(x)]\,dx } \]

Remember

For vertical strips:

\[ \boxed{ \text{Area} = \int (\text{Upper}-\text{Lower})\,dx } \]

Method

Vertical Strips

Width

\(dx\)

Height

\(y_U-y_L\)

Area Element

\(dA=(y_U-y_L)dx\)

\[ \boxed{ A=\int_a^b (y_U-y_L)\,dx } \]

Best used when:

Both curves are naturally expressed as \(y=f(x)\).

Alternative Method

Horizontal Strips

When the curves are naturally expressed as \(x=f(y)\), use horizontal strips.

\[ \boxed{ A= \int_c^d (x_R-x_L)\,dy } \]

Quantity	Horizontal Strip
Thickness	\(dy\)
Length	\(x_R-x_L\)
Area	\((x_R-x_L)dy\)

Shortcut

Symmetry in Area Problems

Symmetry about the y-axis

If the region is symmetric about the \(y\)-axis, calculate one half and multiply by 2.

\[ \boxed{ A=2\int_0^a f(x)\,dx } \]

Even Function

\[ f(-x)=f(x) \] \[ \boxed{ \int_{-a}^{a}f(x)\,dx = 2\int_0^a f(x)\,dx } \]

Odd Function

\[ f(-x)=-f(x) \] \[ \boxed{ \int_{-a}^{a}f(x)\,dx=0 } \]

Exam Tip:

Before evaluating a lengthy integral, inspect the graph for symmetry.

Important Graphs

Standard Regions

Parabola

\[ y=x^2 \]

Circle

\[ x^2+y^2=a^2 \]

\[ A=\pi a^2 \]

Straight Line

\[ y=mx+c \]

Rectangular Hyperbola

\[ xy=c \]

Worked Problems

Solved Examples

Example 1

Area under a parabola

Find the area bounded by \(y=x^2\), \(x=0\), \(x=2\) and the \(x\)-axis.

\[ A=\int_0^2x^2\,dx \] \[ = \left[ \frac{x^3}{3} \right]_0^2 \] \[ = \frac83 \]

\[ \boxed{ A=\frac83 \text{ square units} } \]

Example 2

Area between \(y=x\) and \(y=x^2\)

Find the area enclosed by \[ y=x \] and \[ y=x^2. \]

Step 1: Find intersections

\[ x=x^2 \] \[ x(x-1)=0 \] \[ x=0,\;1 \]

Step 2: Identify upper curve

For \(0\le x\le1\), \[ x>x^2. \]

Step 3: Integrate

\[ A= \int_0^1 (x-x^2)\,dx \] \[ = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 \] \[ = \frac12-\frac13 = \frac16 \]

\[ \boxed{ A=\frac16 \text{ square units} } \]

Example 3

Parabola and line

Find the area enclosed by \[ y=x^2 \] and \[ y=2x. \]

\[ x^2=2x \] \[ x(x-2)=0 \] \[ x=0,\;2 \]

The upper curve is \(y=2x\).

\[ A= \int_0^2 (2x-x^2)\,dx \] \[ = \left[ x^2-\frac{x^3}{3} \right]_0^2 \] \[ = 4-\frac83 \] \[ = \frac43 \]

\[ \boxed{ A=\frac43 \text{ square units} } \]

Example 4

Area under \(4-x^2\)

Find the area enclosed by \[ y=4-x^2 \] and the \(x\)-axis.

\[ 4-x^2=0 \] \[ x=\pm2 \]

The region is symmetric about the \(y\)-axis.

\[ A= 2\int_0^2 (4-x^2)\,dx \] \[ = 2 \left[ 4x-\frac{x^3}{3} \right]_0^2 \] \[ = 2 \left( 8-\frac83 \right) \] \[ = \frac{32}{3} \]

\[ \boxed{ A=\frac{32}{3} \text{ square units} } \]

Example 5

Circle using integration

Find the area of the circle \[ x^2+y^2=a^2. \]

In terms of \(y\),

\[ x=\sqrt{a^2-y^2} \]

Total horizontal width:

\[ 2\sqrt{a^2-y^2} \]

\[ A= \int_{-a}^{a} 2\sqrt{a^2-y^2}\,dy \] \[ = 4\int_0^a \sqrt{a^2-y^2}\,dy \] \[ = \pi a^2 \]

\[ \boxed{ A=\pi a^2 } \]

Higher Order Thinking

HOT Questions

HOT 1

Without directly evaluating separate positive and negative integrals, explain why \[ \int_{-a}^{a}x^3\,dx=0 \] even though the total geometrical area enclosed by \(y=x^3\) and the \(x\)-axis is not zero.

HOT 2

A student wants to find the area between \[ y=x^2 \] and \[ y=4. \] The student uses \[ \int_{-2}^{2} (x^2-4)\,dx. \] Identify the conceptual error and construct the correct integral.

HOT 3

The same region can sometimes be evaluated using \(dx\) or \(dy\). Explain how the choice of strip can change the complexity of the integral without changing the geometrical area.

HOT 4

Find the area enclosed by \[ y=x^2 \] and \[ y=2x+3. \] Your solution must include: intersection points, identification of the upper curve and the definite integral.

HOT 5

A region is symmetric about the \(y\)-axis. One student evaluates the entire integral from \(-a\) to \(a\), while another evaluates only from \(0\) to \(a\) and doubles the result. Explain why both methods should give the same geometrical area when the integrand represents the upper boundary.

Last-Minute Revision

Quick Revision

Quick Revision Checklist

✓ Area bounded by the curve y = f(x), x-axis and ordinates x = a, x = b is A = ∫[a to b] |y| dx.
✓ Area bounded by the curve x = g(y), y-axis and abscissae y = c, y = d is A = ∫[c to d] |x| dy.
✓ Area enclosed between two intersecting curves y = f(x) and y = g(x) is A = ∫[a to b] |f(x) - g(x)| dx.
✓ Standard symmetrical regions (circles, ellipses, parabolas) can be solved by computing the first quadrant area and multiplying by symmetry factor.