Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Area bounded by the curves <img alt="y = \sin \frac{{\pi x}}{2}$" src="https://learn.careers36

Area bounded by the curves y = \sin \frac{{\pi x}}{2}$ and y = {x^3}$, is equal to

Option: 1

\frac{{4 - \pi }}{\pi }sq. units


Option: 2

\frac{{4 - \pi }}{{2\pi }}$sq. units


Option: 3

\frac{{8 - \pi }}{\pi }$sq. units


Option: 4

\frac{{8 - \pi }}{{2\pi }}$sq. units


Answers (1)

best_answer

As we learnt 

Area between two curves -

\\*I\! \! f \: f\left ( x \right )\geqslant g\left ( x \right )\\* in[a,c)\: \: and \: \: g\left ( x \right ) \geqslant f\left ( x \right )\:in(c,b]\\* Then\: area = \\*\\* \int_{a}^{c}\left ( f\left ( x \right )-g\left ( x \right ) \right )dx+\int_{c}^{b}\left ( g\left ( x \right )-f\left ( x \right ) \right )dx

- wherein

 

 

 

 

Bounded  area D =

2\int\limits_0^1 {\left( {\sin \frac{{\pi x}}{2} - {x^3}} \right)} \,dx$

=2\left( { - \frac{2}{\pi }\cos \frac{{\pi x}}{2} - \frac{{{x^4}}}{4}} \right)_0^1 = 2\left( {\frac{2}{\pi } - \frac{1}{4}} \right) = \frac{{8 - \pi }}{{2\pi }}$

Posted by

mansi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions