Applications of the Integrals Class 12 Notes Maths Chapter 8 - CBSE

Chapter:8

What are Applications of the Integrals ?

https://drive.google.com/file/d/1CvWpyW73lM0F-eJt07Mvd-C2HDIKzzy7/view

Applications of the integrals means ‘Use of integrals’.
For finding the area of any curve we must have knowledge of integrals and conic section.

Area Under Simple Curve

We consider the easy and intuitive way to finding the area bounded by the curve. From the figure ,we think of area under the curve as composed of large number of very thin vertical strip (black color strip). And we consider this black vertical strip as a rectangle .

Case 1:

The area bounded by the curve y = f(x), the x-axis and the ordinates
x = a and x = b is given by (above figure) As we know area of rectangle ‘A’= length x breath

dA = y dx
⇒ dA = f(x)dx

$$\text{Integrate both sides}\\ \implies\int^b_a\text{dA}=\int^b_a\text{f(x)dx}\\ \implies \text{A}=\int^b_a\text{f(x)dx}$$

Case 2:

The area bounded by the curve x = f(y), the y-axis and the abscissa
y = c and y = d is given by Similarly, in case 1

$$\text{A}=\int^d_c\text{f(y)dy}$$

Case 3:

If the curve y = f(x) lies below the x−axis, then the area bounded by the curve y = f(x), the x−axis and the lines x = a and x = b come out to be negative. Thus , if the area is negative (Area never negative), we take its absolute value, i.e.;

$$\text{A} = |\int^b_a \text{f(x)dx}|$$

Area Enclosed By The Parabola (X² = 4ay) And Its Latus Rectum

In Case Of Horizontal Strip:

$$\text{A} = 4\sqrt{a} \int^a_0 \sqrt \text{y\space dy}\\\implies\text{A} = 4\sqrt{a}\bigg[\frac{\text{y}^{\frac{3}{2}}}{\frac{3}{2}}\bigg]=4\sqrt{a}×\frac{2}{3}a^{\frac{3}{2}}\\\implies \text{A}=\frac{8a^2}{3}\\\text{We get same answer in case of vertical strip i.e A=}\frac{8a^2}{3}$$

Area Enclosed By The Ellipse (x²/a²+y²/b²=1)

In Case Of Horizontal Strip:

$$\text{A} = 4\frac{a}{b}\int^b_0 \sqrt{b^2-y^2}\text{dy}\\ \implies\text{A}=\pi\text{ab}\\\text{We get same answer in case of vertical strip i.e. A }= \pi\text{ab}\space \\(\because \text{A}=4\frac{b}{a}\int^a_0\sqrt {a^2-x^2} \space\text{dx})$$

Area Enclosed By The Circle (X² + Y² = a²)

In Case Of Horizontal Strip:

$$\text{A} = 4\int^a_0 \sqrt{a^2-y^2}\text{dy}\\\qquad\text{A}=\pi a^2\\ \text{We get same answer in case of vertical strip i.e. A = }\pi a^2\\ (\therefore \text{A}=4\int^a_0\sqrt {a^2-x^2}\space \text{dx}) $$