We want to compute the definite integral of the function $f(x) = 8x^3 \sqrt{x^2 + 4}$ from 0 to 2.

The integral is written as:

$I = \int_{0}^{2} 8x^3 \sqrt{x^2 + 4} \, dx$

Step 1: Substitution

Let’s use substitution to simplify the integral. Set:

$u = x^2 + 4$

Then,

$du = 2x \, dx \quad \text{or} \quad \frac{du}{2} = x \, dx$

Also, since $u = x^2 + 4$, we have:

• When $x = 0$, $u = 4$,
• When $x = 2$, $u = 8$.

Step 2: Rewrite the Integral

Now, rewrite the original function in terms of $u$. Since $x^2 = u - 4$, we have:

$x^3 = x \cdot x^2 = x \cdot (u - 4)$

So, the integral becomes:

$I = \int_{4}^{8} 8(u - 4) \sqrt{u} \cdot \frac{du}{2}$

Simplifying:

$I = 4 \int_{4}^{8} (u - 4) \sqrt{u} \, du$

Step 3: Break into Two Integrals

Distribute the $\sqrt{u}$ term:

$I = 4 \left( \int_{4}^{8} u^{3/2} \, du - 4 \int_{4}^{8} u^{1/2} \, du \right)$

Step 4: Compute the Integrals

Now we compute each integral:

1. For $\int u^{3/2} \, du$, use the power rule for integration:

$\int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$

2. For $\int u^{1/2} \, du$, use the same power rule:

$\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

Step 5: Evaluate the Integrals

Now, substitute these results back into the expression for $I$:

$I = 4 \left( \frac{2}{5} u^{5/2} \Big|_{4}^{8} - 4 \cdot \frac{2}{3} u^{3/2} \Big|_{4}^{8} \right)$

1. First, evaluate $u^{5/2}$ at the limits:

$\frac{2}{5} \left( 8^{5/2} - 4^{5/2} \right) = \frac{2}{5} \left( 32\sqrt{8} - 32 \right)$

2. Then evaluate $u^{3/2}$ at the limits similarly:

Complete