#### To determine

**a) $2:** ex≥1+x+x22**if** x≥0

#### Answer

**Solution**: ex≥1+x+x22**if** x≥0

#### Explanation

To prove this we will instead show f(x)=ex−(1+x+x22) is always positive

Let f(x)=ex−(1+x+x22)

We have f(0)=e0−1−0=1−1=0

Differntiating with respect to x

f′(x)=ex−1−x

Clearly f′(x)=ex−1−x>0 ∀x≥0

Since f(x)=ex−(1+x+x22) is increasing on (0,∞)

Therefore f(x)=ex−(1+x+x22)≥0⇒ex≥1+x+x22

**Conclusion: **ex≥1+x+x22**if** x≥0

#### To determine

b) **To improve:** the estimate 43≤∫01ex2dx≤e using ex≥1+x+x22

#### Answer

43≤∫01ex2dx≤e

#### Explanation

**Explanation: ** We know that 1+x2+x42≤ex2≤ex

Integrating with respect to x from x=0 to x=1, we get

∫01(1+x2+x42)dx≤∫01ex2dx≤∫01exdx⇒x+x33+x510|01≤∫01ex2dx≤ex|01⇒1+13+110≤∫01ex2dx≤e−1⇒4330≤∫01ex2dx≤e−1

**Conclusion: **43≤∫01ex2dx≤e