If the area bounded by the curve y=2kx,k>0 and x=0, x=2 and x− axis is 3ln2, then the value of k is
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Solution
The given curve is y=2kx,k>0
Clearly, y>0 for all values of x and k ∴ the curve y=22k lies above x−axis for 0≤x≤2.
Area bounded by the curve y=2kx,x=0,x=2 is given by: A=2∫0ydx=2∫02kxdx =1k2k∫02tdt, where t=kx =1k[2tln2]2k0=1kln2(22k−1)
Now, A=3ln2(given) ∴1kln2(22k−1)=3ln2⇒22k−1=3k ⇒k=1