#### To determine

**To find:**

The values of m so that the line y=mx and the curve y=xx2+1 enclose a region and find the area of the region

#### Answer

The curve and the line enclose a region when 0<m<1

Area = ln1m-1+m

#### Explanation

**1) Concept:**

Area is the integral of the difference of two functions.

2) **Calculation:**

Given that

y=mx, y=xx2+1

To find the intersection points of the given curves, equate both to each other.

mx=xx2+1

Multiply by x2+1

mx·x2+1=xx2+1·x2+1

By simplifying,

mx3+mx=x

Subtract by x.

mx3+mx-x=0

Simplify.

xmx2+m-1

Simplify.

x=0 or mx2+m-1

Solve for x by quadratic formula.

x=±1m-1

For 0<m<1, the intersection of two functions will be at x=0, and at x=±k where k=1m-1.

Since the region is symmetric, there will be two equal areas.

Area =∫-kkxx2+1-mxdx

By using ∫-aaf(x)dx=2∫0af(x)dx

A=2∫0kxx2+1-mxdx

By solving the integral,

A=lnx2+1-mx20k

By substituting limits,

=lnk2+1-mk2-ln02+1-m0

By using k=1m-1

=ln1m-12+1-m1m-12

Simplify.

ln1m-1+1-m1m-1

Simplify.

ln1m-1+m

Therefore, the values of m so that the line y=mx and the curve y=xx2+1 enclose a region bounded 0<m<1and area is ln1m-1+m

**Conclusion:**

The enclosed region is 0<m<1

Area = ln1m-1+m