Area Under The Curve Using Integration | Class 12 Math Notes Study Material Download Free PDF

How to Determine the Area Under the Curve?

Let us assume the curve y=f(x) and its ordinates at the x-axis be x=a and x=b. Now, we need to evaluate the area bounded by the given curve and the ordinates given by x=a and x=b.

Area under the Curve

The area under the curve can be assumed to be made up of many vertical, extremely thin strips. Let us take a random strip of height y and width dx as shown in the figure given above whose area is given by dA.

The area dA of the strip can be given as y dx. Also, we know that any point of the curve, y is represented as f(x). This area of the strip is called an elementary area. This strip is located somewhere between x=a and x=b, between the x-axis and the curve. Now, if we need to find the total area bounded by the curve and the x-axis, between x=a and x=b, then it can be considered to be made of an infinite number of such strips, starting from x=a to x=b. In other words, adding the elementary areas between the thin strips in the region PQRSP will give the total area.

Mathematically, it can be represented as:

\(\begin{array}{l} A = \int\limits_{a}^b dA = \int\limits_{a}^b ydx = \int\limits_{a}^b f(x)dx\end{array} \)

Using the same logic, if we want to calculate the area under the curve x=g(y), y-axis between the lines y=c and y=d, it will be given by:

Area Under The Curve -2

\(\begin{array}{l} A = \int\limits_{c}^d xdy = \int\limits_{c}^d g(y)dy\end{array} \)

In this case, we need to consider horizontal strips as shown in the figure above.

Also, note that if the curve lies below the x-axis, i.e. f(x) <0 then following the same steps, you will get the area under the curve and x-axis between x=a and x=b as a negative value. In such cases, take the absolute value of the area, without the sign, i.e.|

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\(\begin{array}{l}\int\limits_{a}^b f(x)dx|\end{array} \)

Area Under The Curve -3

Another possibility is that, when some portion of the curve may lie above the x-axis and some portion below the x-axis, as shown in the figure,

Area under the Curve -4

Here A1<0 and A2>0. Hence, this is the combination of the first and second case. Hence, the total area will be given as |A1|+A2

Area Under the Curve Example

Let us consider an example, to understand the concept in a better way.

We need to find the total area enclosed by the circle x2+y2=1

Area Under The Curve Example

Area enclosed by the whole circle = 4 x area enclosed OABO

=4

\(\begin{array}{l}\int\limits_{0}^1 ydx  \end{array} \)

(considering vertical strips)

\(\begin{array}{l} =4 \int\limits_{0}^1 \sqrt{{1}-{x}^{2}}\end{array} \)

On integrating, we get,

\(\begin{array}{l}=4 [\frac{x}{2}\sqrt{{1}-{x}^{2}}+ \frac{1}{2} {sin}^{-1}x{]}_{0}^{1}\end{array} \)

= 4 × 1/2 x π/2

= π

So the required area is π square units.

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