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Home / Doubts and Solutions/JEE/Physics

Using Definite Integration :

qsnImg
Asked by Anish 25th July 2018, 11:53 AM
Answered by Expert
Answer:
As shown in figure, we need to find the enclosed area between the curve x2 + y2 = 4 and the straight line x+y=2
 
begin mathsize 12px style A r e a space equals space integral subscript 0 superscript 2 square root of 4 minus x squared end root space d x space space minus space integral subscript 0 superscript 2 space left parenthesis 2 minus x right parenthesis space d x space space equals space I subscript 1 space minus space I subscript 2 space space end style
begin mathsize 12px style w h e r e space I subscript 1 equals integral subscript 0 superscript 2 square root of 4 minus x squared end root space d x space space space a n d space space I subscript 2 space equals space integral subscript 0 superscript 2 left parenthesis 2 minus x right parenthesis d x F i r s t space i n t e g r a l space i s space s o l v e d space u sin g space t h e space s u b s t i t u t i o n space x space equals space 2 cos theta I subscript 1 space equals integral subscript 0 superscript bevelled pi over 2 end superscript 4 sin squared theta space d theta space equals space 2 integral subscript 0 superscript bevelled pi over 2 end superscript left parenthesis 1 minus cos space 2 theta right parenthesis d theta space equals pi I subscript 2 space equals integral subscript 0 superscript 2 left parenthesis 2 minus x right parenthesis d x space equals space 2 A r e a space equals space I subscript 1 minus I subscript 2 space equals left parenthesis pi minus 2 right parenthesis space s q. u n i t s end style
Answered by Expert 25th July 2018, 2:26 PM
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