integrate between 0 and 2 the function (8x^3)*(sqrt(x^2+4))

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

Learn how to compute the definite integral of the function 8x^3 * sqrt(x^2 + 4) from 0 to 2 using substitution and integration techniques. This step-by-step solution covers key concepts like substitution, power rule, and the Fundamental Theorem of Calculus. Suitable for undergraduate calculus students.

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