Calculate the integral: integral from 0 to 4 x³

Calculate the integral: $\int_0^4 x^3 \sqrt{x^2+9} \, dx$.

We will use substitution to evaluate this integral.

Step 1: Choose a substitution

Let $u = x^2 + 9$. Then $du = 2x \, dx$, so $x \, dx = \frac{1}{2} du$, and $x^2 = u - 9$.

Step 2: Rewrite the integrand

\[\int_0^4 x^3 \sqrt{x^2+9} \, dx = \int_0^4 x^2 \cdot x\sqrt{x^2+9} \, dx\]

Substituting $x^2 = u - 9$ and $x\sqrt{x^2+9} \, dx = \sqrt{u} \cdot \frac{1}{2} du$:

\[= \int (u-9) \cdot \sqrt{u} \cdot \frac{1}{2} \, du\]

Step 3: Change the limits of integration

When $x = 0$: $u = 9$. When $x = 4$: $u = 25$.

Step 4: Rewrite the integral with new limits

\[\int_0^4 x^3 \sqrt{x^2+9} \, dx = \frac{1}{2}\int_9^{25} (u-9)u^{1/2} \, du = \frac{1}{2}\int_9^{25} (u^{3/2} - 9u^{1/2}) \, du\]

Step 5: Integrate term by term

\[\int u^{3/2} \, du = \frac{2u^{5/2}}{5}, \qquad \int u^{1/2} \, du = \frac{2u^{3/2}}{3}\]

Therefore:

\[= \frac{1}{2}\left[\frac{2u^{5/2}}{5} - 6u^{3/2}\right]_9^{25} = \left[\frac{u^{5/2}}{5} - 3u^{3/2}\right]_9^{25}\]

Step 6: Evaluate at the upper limit $u = 25$

$25^{5/2} = 3125$, $25^{3/2} = 125$:

\[F(25) = \frac{3125}{5} - 3(125) = 625 - 375 = 250\]

Step 7: Evaluate at the lower limit $u = 9$

$9^{5/2} = 243$, $9^{3/2} = 27$:

\[F(9) = \frac{243}{5} - 3(27) = \frac{243}{5} - \frac{405}{5} = \frac{-162}{5}\]

Step 8: Calculate the final result

\[\int_0^4 x^3 \sqrt{x^2+9} \, dx = F(25) - F(9) = 250 + \frac{162}{5} = \frac{1250}{5} + \frac{162}{5} = \frac{1412}{5}\]

Verification using the antiderivative:

The antiderivative of $x^3\sqrt{x^2+9}$ is:

\[F(x) = \frac{(x^2+9)^{5/2}}{5} - 3(x^2+9)^{3/2}\]

Differentiating: $F'(x) = x(x^2+9)^{3/2} - 9x(x^2+9)^{1/2} = x(x^2+9)^{1/2}\cdot x^2 = x^3\sqrt{x^2+9}$

Geometric interpretation:

The integral $\int_0^4 x^3 \sqrt{x^2+9} \, dx$ represents the area under the curve $y = x^3\sqrt{x^2+9}$ from $x = 0$ to $x = 4$. At $x=0$: $y=0$; at $x=4$: $y=4^3\sqrt{25}=64\cdot5=320$. The function grows rapidly since it involves both $x^3$ and $\sqrt{x^2+9}$, and is always positive and increasing on $[0,4]$.

Numerical value:

\[\frac{1412}{5} = 282.4\]

Final answer:

\[\boxed{\int_0^4 x^3 \sqrt{x^2+9} \, dx = \frac{1412}{5}}\]

Or in mixed number form: $282\frac{2}{5}$ or as a decimal: $282.4$