Find the area bounded by the x-axis, part of the curve `y=(1-8/(x^2))` , and the ordinates at `x=2a n dx=4.` If the ordinate at `x=a` divides the area into two equal parts, then find `adot`

Answer

`4 " sq. units ; "a=2sqrt(2)`

Explanation

`"Given "y=1+(8)/(x^(2))`.
Here y is always positive, hence, curve is lying above the a-axis.
`therefore" Req. area "=int_(2)^(4)ydx=int_(2)^(4)(1+(8)/(x^(2)))dx`
`=[x-(8)/(x)]_(2)^(4)=4.`
If x=a bisects the area, then we have
`overset(a)underset(2)int(1+(8)/(x^(2)))dx=[x-(8)/(x)]_(2)^(a)=[a-(8)/(a)-2+4]=(4)/(2).`
`"or "a-(8)/(a)=0`
`"or "a^(2)=8`
`"or "a=pm2sqrt(2)`
`"Since "agt2,a=2sqrt(2).`