#### To determine

**To find:**

the values of m such that the curves y=mx and y=xx2+1 enclose the region and hence find the area

#### Answer

A=12(lnm−1+m)

#### Explanation

**Given:**

y=mx and y=xx2+1

The point of intersection of the curves is given by

mx=xx2+1⇒m=1x2+1clearly,0<m<1

The limit of integration are the points of intersection of both curves y=mx and y=xx2+1

mx=xx2+1⇒x(m−1x2+1)=0⇒x=0 or m=1x2+1

m=1x2+1⇒x2+1=1m⇒x2=1m−1⇒x=(1m−1) Area enclosed by the curves is given by

A=∫01m−1[xx2+1−mx] dx

The limits of x are obtained by

m=1x2+1⇒x2+1=1m⇒x2=1m−1⇒x=1m−1

Therefore, we have

A=∫01m−1[xx2+1−mx] dx=12ln(x2+1)−mx22|01m−1=12(lnm−1+m)

**Conclusion:**

A=12(lnm−1+m)