Home | | **Business Maths 12th Std** | Geometrical Interpretation of Definite Integral as Area under a curve

Using integration we can evaluate the area bounded by the curves with coordinate axes.

**The area of the region bounded by the curves**

Using integration we can evaluate the area bounded by the curves
with coordinate axes. We can also calculate the area between two given curves.

**Geometrical Interpretation of Definite
Integral as Area under a curve:**

Suppose we want to find out the area of the region which is
bounded above by a curve *y* =
*f* ( *x*) , below by the *x*
âˆ’ *axis* and the
lines *x* =
*a* and *x* =
*b* .

Now from Fig 3.1 let the interval [a b] is divided into n
subintervals [x_{i-1}, x_{i}] of equal length âˆ†*x*_{i} i.e x_{i}-x_{i-1}
= âˆ†_{xi} for any x_{i}â€™ âˆˆ [x_{i-1},x_{i}], let f(xâ€²_{i})
be the height of n rectangles having x_{i}-x_{i-1}
= âˆ†x_{i} as its base. Then area A_{i} = âˆ†x_{i} *f* (x_{i}â€™). Now the total area A
=

Now from the definition of definite integral, if *f* ( *x*) is a function defined on
[ *a* , *b*] with *a* < *b*
then the definite integral is

The area under the curve is exhausted by increasing the number of
rectangular strips to âˆž

Thus the geometrical interpretation of definite integral is the
area under the curve between the given limits.

The area of the region bounded by the curve *y*=*f*(*x*),
with *x*- axis and the ordinates at *x *=* a *and* x *=* b *given by

**Note **

(i) The area of the region bounded by the curve y = f (x)
between the limits x = a , x = b and lies below x -axis, is

(ii) The area of the
region bounded by the curve x = f (y) between the limits y = c and y = d with y
âˆ’ axis and the area lies lies to the right
of y- axis, is

(iii) The area bounded by the curve *x* = *f* (*y*)
between the limits *y *=* c *and* y *=* d with y *âˆ’* axis *and the area lies to the left* *of *y*- axis, is

Let *f*(*x*) and *g*(*x*) be two continuous
functions defined on *x* in the interval [*a*, *b*]. Also *f*
(*x*) > *g* (*x*),*a*
â‰¤ *b*

Then the area between these two curves from *x* = *a to x* = *b* , is

**Example 3.1**

Find the area bounded by *y* = 4*x* +
3 with *x*- axis between the lines *x* = 1 and *x* = 4* *

*Solution:*

**Example 3.2**

Find the area of the region bounded by the line *x *âˆ’* *2* y *âˆ’12* *=* *0* *, the* y*-axis
and the lines* y *= 2,* y *= 5.

*Solution:*

*x *âˆ’* *2* y *âˆ’12*
*=* *0

*x *=* *2* y *+* *12

Required Area

**Example 3.3**

Find the area of the region bounded by the parabola *y* = 4 âˆ’ *x*^{2} , *x*
âˆ’ *axis* and the
lines *x *=* *0,* x *=* *2

*Solutions:*

**Example 3.4**

Find the area bounded by *y* = *x* between the lines *x* = âˆ’1*and x* = 2 with *x* -axis.

*Solutions:*

**Example 3.5**

Find the area of the parabola *y* ^{2} = 8*x* bounded by
its latus rectum.

*Solution*

y^{2}=8x
(1)

Comparing this with the standard form *y* ^{2} = 4*ax* ,

4*a* =
8

*a *=* *2

Equation of latus rectum is *x* = 2

Since equation (1) is symmetrical about *x*- axis

Required Area = 2[Area in the first quadrant between the limits *x*
= 0 and *x* = 2]

**Example 3.6**

Sketch the graph *y* =
|*x* + 3| and evaluate

*Solution:*

**Example 3.7**

Using integration find the area of the circle whose center is at
the origin and the radius is a units.

*Solution*

Equation
of the required circle is x^{2} + y^{2} = a^{2} (1)

put
y = 0, x^{2} = a^{2}

â‡’ x = Â± a

Since
equation (1) is symmetrical about both the axes

The
required area = 4 [Area in the first quadrant between the limit 0 and a.]

**Example 3.8**

Using integration find the area of the region bounded between the
line *x* = 4 and the parabola *y*
^{2} = 16*x*.

*Solution:*

The equation *y* ^{2} = 16*x* represents a parabola (Open
rightward)

Tags : Integral Calculus - II , 12th Business Maths and Statistics : Chapter 3 : Integral Calculus - II

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12th Business Maths and Statistics : Chapter 3 : Integral Calculus - II : Geometrical Interpretation of Definite Integral as Area under a curve | Integral Calculus - II

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