Calculate the integral: integral from 0 to 4 x sqrt(x²+9)

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

We will use substitution to evaluate this integral.

Step 1: Choose a substitution

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

Step 2: Change the limits of integration

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

Step 3: Rewrite the integral

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

Step 4: Integrate using the power rule

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

Step 5: Evaluate at the limits

At $u = 25$: $25^{3/2} = 125$. At $u = 9$: $9^{3/2} = 27$.

Step 6: Calculate the final result

\[\int_0^4 x \sqrt{x^2+9} \, dx = \frac{1}{3}[125 - 27] = \frac{98}{3}\]

Verification using the antiderivative:

If $F(x) = \frac{1}{3}(x^2+9)^{3/2}$, then $F'(x) = x\sqrt{x^2+9}$. This confirms our substitution was correct!

\[\int_0^4 x \sqrt{x^2+9} \, dx = \left[\frac{1}{3}(x^2+9)^{3/2}\right]_0^4 = \frac{125}{3} - \frac{27}{3} = \frac{98}{3}\]

Geometric and physical interpretation:

• Mathematically, the integral represents the area under the curve $y = x\sqrt{x^2+9}$ from $x = 0$ to $x = 4$.
• At $x = 0$: $y = 0$. At $x = 4$: $y = 4\sqrt{25}=20$.
• The function is continuously increasing on $[0, 4]$.
• The expression $\sqrt{x^2+9}$ represents the distance from the point $(x, 0)$ to the point $(0, 3)$ in the plane.

Pattern recognition:

This integral follows a common pattern for $\int x\sqrt{ax^2+b} \, dx$: the presence of $x$ outside the square root suggests $u=x^2+b$, converting the integral into a simple power function, with antiderivative $\frac{1}{3a}(ax^2+b)^{3/2}+C$.

Numerical verification:

\[\frac{98}{3} = 32.\overline{6} \approx 32.667\]

Final answer:

\[\boxed{\int_0^4 x \sqrt{x^2+9} \, dx = \frac{98}{3}}\]

Or in mixed number form: $32\frac{2}{3}$