#### To determine

**To show:**

0≤∫510x2x4+x2+1dx≤0.1

#### Answer

0≤∫510x2x4+x2+1dx≤0.1

#### Explanation

**1) Concept:**

Use fundamental theorem of calculus

**2) Calculations:**

Since 0≤x2 and 0≤x4 for all x∈R,

0≤x2+x4

Adding 1 to both sides of the above inequality,

1≤1+x2+x4

Also, 0≤x4≤x4+x2+1 for all x∈R.

Hence, divide x4+x2+1 to the above inequality

0≤x4x4+x2+1≤1 for all x∈R.

Since x2>0, for all x∈5, 10, divide x2 to the above inequality

0≤x2x4+x2+1≤1x2 for all x∈5, 10.

Integrating between limits 5 to 10

∫5100dx≤∫510x2x4+x2+1 dx≤∫5101x2 dx Ant derivate of 1x2 is -1x. Thus using fundamental theorem of calculus,

0105≤∫510x2x4+x2+1 dx≤-1x105

That is 0≤∫510x2x4+x2+1 dx≤-110--15

After simplifying,

0≤∫510x2x4+x2+1dx≤110

**Conclusion:**

Therefore,0≤∫510x2x4+x2+1dx≤0.1