Important Note

Please turn desktop mode or rotate your mobile screen for better view

Definitions and Formulas

Application of Integrals:- Application of Integrals is finding the area of curves.

Area Under Simple Curves:-

(1) The area A of the region bounded by the curve y = f(x), the x-axis and line x = a, x = b is given by:-

{"type":"align*","id":"1-0","code":"\\begin{align*}\n{A}&={\\int_{a}^{b}dA}\\\\\n{A}&={\\int_{a}^{b}f\\left(x\\right).dx}\\\\\n{A}&={\\int_{a}^{b}y.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1603271903317,"cs":"jkmGLdY7qQZxFHjGEwMvpw==","size":{"width":112,"height":126}}

But Below x-axis:-

{"id":"1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{A}&={\\left|\\int_{a}^{b}dA\\right|}\\\\\n{A}&={\\left|\\int_{a}^{b}f\\left(x\\right).dx\\right|}\\\\\n{A}&={\\left|\\int_{a}^{b}y.dx\\right|}\t\n\\end{align*}","ts":1603272245913,"cs":"q5nco0uQl/lYKM4StAlfnQ==","size":{"width":121,"height":136}}

(2) The area A of the region bounded by the curve x = g(y), the y-axis and line y = c, y = d is given by:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"1-1-0","code":"\\begin{align*}\n{A}&={\\int_{c}^{d}dA}\\\\\n{A}&={\\int_{c}^{d}g\\left(y\\right).dy}\\\\\n{A}&={\\int_{c}^{d}x.dy}\t\n\\end{align*}","ts":1603271964547,"cs":"iiuXvAoMaWx8b2dHCC5qqw==","size":{"width":108,"height":126}}

But in -y-axis:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"1-1-1","code":"\\begin{align*}\n{A}&={\\left|\\int_{c}^{d}dA\\right|}\\\\\n{A}&={\\left|\\int_{c}^{d}g\\left(y\\right).dy\\right|}\\\\\n{A}&={\\left|\\int_{c}^{d}x.dy\\right|}\t\n\\end{align*}","type":"align*","ts":1603272277603,"cs":"bFA5huktKkcU3o3N85qXbg==","size":{"width":120,"height":136}}

Area between two curves:-

If y = f(x) and y = g(x) 

where f(x) > g(x) in [a,b] 

then area between curves f(x) and g(x)

{"id":"1-0","type":"align*","code":"\\begin{align*}\n{A}&={\\int_{a}^{b}dA}\\\\\n{A}&={\\int_{a}^{b}\\left[f\\left(x\\right)-g\\left(x\\right)\\right].dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603273433047,"cs":"UT7vwKTxtTZnThlQr2+LTQ==","size":{"width":165,"height":82}}

Or A = [area bounded by y = f(x), x-axis  and line x = a, x = b]

- [area bounded by y = g(x), x-axis and line x = a, x = b]

{"id":"1-0","code":"\\begin{align*}\n{A}&={\\int_{a}^{b}f\\left(x\\right).dx-\\int_{a}^{b}g\\left(x\\right).dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1603273621129,"cs":"+ZJDPIEOBUF/y627vcc2mA==","size":{"width":208,"height":38}}

where f(x) > g(x)

Example:- Find the area enclosed by the circle x2 + y2 = a2.

Solun:- Area of given circle is:- 4(area of the region AOBA)........(1)

Area of the region AOBA:-

{"code":"\\begin{align*}\n{}&={\\int_{0}^{a}y.dx}\\\\\n{}&={\\int_{0}^{a}{\\sqrt[]{a^{2}-x^{2}}}.dx}\t\n\\end{align*}","id":"1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1603276248178,"cs":"MPolOy0PwnI8VQl5C/tWqg==","size":{"width":136,"height":77}}

We know that:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"2","ts":1603276347627,"cs":"9Ycqurei9GIIzBsh9pvVFg==","size":{"width":332,"height":20}}

{"type":"align*","code":"\\begin{align*}\n{}&={\\left[\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)\\right]_{0}^{a}}\\\\\n{}&={\\left[\\left(\\frac{a}{2}{\\sqrt[]{a^{2}-a^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{a}{a}\\right)\\right)-\\left(\\frac{0}{2}{\\sqrt[]{a^{2}-0}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{0}{a}\\right)\\right)\\right]}\\\\\n{}&={\\left(\\frac{a}{2}\\times0+\\frac{a^{2}}{2}\\sin^{-1}\\left(1\\right)\\right)-0}\\\\\n{}&={\\frac{a^{2}}{2}\\times\\frac{\\Pi}{2}}\\\\\n{}&={\\frac{\\Pi a^{2}}{4}}\t\n\\end{align*}","id":"3","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603276630110,"cs":"dkX3DkHMj2T79aiS7Casxw==","size":{"width":486,"height":206}}

Area of given circle is = 4(area of the region AOBA) = 4(Ï€a2/4)

Area of the given circle is = πa2.


If you have any queries, you can ask me in the comment section

And you can follow/subscribe to me for the latest updates on your e-mails

For subscribing me follow these instructions:-
1. Fill your E-mail address
2. Submit Recaptcha
3. Go to your email and then click on the verify link
Then you get all update on your email

Thanks for Reading ......
Tags

Post a Comment

Comment me for any queries or topic which you want to learn

Post a Comment

Comment me for any queries or topic which you want to learn

Previous Post Next Post