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Question

Mathematics Question on applications of integrals

Find the area bounded by the curve y=sin x between x=0 and x=2π

Answer

The graph of y=sin x can be drawn as

∴Required area=Area OABO+Area BCDB

=

0πsinxdx\int_{0}^{π} \sin x \,dx

+π2πsinxdx+\int_{π}^{2π} \sin x\,dx

=[-cosx]π0+|[-cosx]2ππ|

=[-cosπ-cos0]+|-cos2π+cosπ|

=1+1+|(-1-1)|

=2+|-2|

=2+2=4units