Answer
$\int_{1}^{3} \sqrt{x^4+1}~dx \geq \frac{26}{3}$
Work Step by Step
According to Exercise 28:
$\int_{1}^{3} x^2~dx = \frac{3^3-1^3}{3} = \frac{26}{3}$
On the interval $1 \leq x \leq 3$:
$x^4+1 \geq x^4$
$\sqrt{x^4+1} \geq \sqrt{x^4} = x^2$
Therefore, by Property 7:
$\int_{1}^{3} \sqrt{x^4+1}~dx \geq \int_{1}^{3} x^2~dx = \frac{26}{3}$
$\int_{1}^{3} \sqrt{x^4+1}~dx \geq \frac{26}{3}$
