Question

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

Answer 1

`I=8pi`

Explanation:

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`