Question

Area bounded by the curves $\frac{y}{x}=\log x$ and $\frac{y}{2}=-x^2+x$ equals:

• A. 7/12
• B. 12/7
• C. 7/6
• D. 6/7

Answer 1

Answer: (A)

7/12

The two curves intersect at $x=0$ and $x=1$ can be found graphically.
So area bound= ${\int }_0^1(-2x^2+2x-xlogx)dx$

$=(\frac{-2x^3}{3}+x^2){|}_0^1-\int xlogxdx$
to Integrate $xlogxdx$, substitute $logx=t,$
$dx=xdt$
Integral becomes $\int e^{2t}tdt$
Integration by parts and then putting limits gives $=-1/4$

So bound area $=-2/3+1+1/4$
$=7/12$