Integral help how do I do this one

int 2L *sqrt(r^2 -(x-r)^2) dx

from 0 to 2r

I = 2L ∫ [ 0, 2r ] √[ r² - (x-r)² ] dx ......... (1)
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Let : u = x - r. ∴ du = dx.

Also :

x = 0 ⇒ u = 0 - r = -r

x = 2r ⇒ u = 2r - r = r.
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From (1), then,

I = 2L ∫ [ -r, r ] √( r² - u² ) du

= 2L • 2 ∫ [ 0, r ] √( r² - u² ) du

= 2L • 2 • (1/2) [ u.√(r²-u²) + r². sin ֿ¹ (u/r) ] ... on [ 0, r ]

= 2L • [ ( 0 + r². sin ֿ¹ (1) ) - ( 0 + r². sin ֿ¹ (0) ) ]

= 2L • [ r². ( π/2 ) - (0) ]

= π r² L ........................................… Ans.
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Happy To Help !
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