What is #int_(0)^(8) sqrt( 8x-x^(2))dx #?

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Answer 1 · 2018-04-18T08:43:00

#I=8pi#

Here,

#I=int_0^8 sqrt(8x-x^2dx#

#=int_0^8sqrt(16-(x^2-8x+16))dx#

#=int_0^8sqrt(4^2-(x-4)^2)dx#

We know that ,

#color(red)(intsqrt(a^2-X^2)dX=X/2sqrt(a^2-X^2)+a^2/2sin^-1(X/a)+c#

Put, #a=4 and X=x-4#

#I=[(x-4)/4sqrt(4^2-(x-4)^2)+4^2/2sin^-1((x-4)/4)]_0^8#

#=>I=[(x-4)/4sqrt(8x-x^2)+8sin^-1((x-4)/4)]_0^8#

#=>I=[(8-4)/4sqrt(64-64)+8sin^-1((8-4)/4)]-[(0-4)/4sqrt(0- 0^2)+8sin^-1((0-4)/4)]#

#=[0+8sin^-1(1)]-[0+8sin^-1(-1)]#

#=8sin^-1(1)-8sin^-1(-1)#

#=8sin^-1(1)+8sin^-1(1)#

#=8(pi/2)+8(pi/2)#

#=4pi+4pi#

#=8pi#