2\pi \int _0^3\left(t+3\right)\sqrt{\left(\frac{t}{2}\right)^2+\left(1\right)^2}dt

To solve the integral 

\[ 2\pi \int_0^3 \left(t + 3\right) \sqrt{\left(\frac{t}{2}\right)^2 + 1} \, dt, \]

we can proceed as follows:

1. **Simplify the integrand:**

   The integrand is \((t + 3) \sqrt{\left(\frac{t}{2}\right)^2 + 1}\).

   Let's simplify the expression inside the square root:
   \[
   \sqrt{\left(\frac{t}{2}\right)^2 + 1} = \sqrt{\frac{t^2}{4} + 1}.
   \]

2. **Substitute:**

   Let \( u = \frac{t^2}{4} + 1 \). Then, we differentiate \( u \) with respect to \( t \):
   \[
   \frac{du}{dt} = \frac{d}{dt} \left( \frac{t^2}{4} + 1 \right) = \frac{t}{2}.
   \]
   Hence,
   \[
   du = \frac{t}{2} \, dt \quad \Rightarrow \quad dt = \frac{2 \, du}{t}.
   \]

   However, directly substituting \( u \) might complicate the integral further. Instead, we consider solving it directly by parts or recognizing a pattern.

3. **Split the integral:**

   Split the integral into two parts to handle separately:
   \[
   2\pi \int_0^3 (t + 3) \sqrt{\frac{t^2}{4} + 1} \, dt = 2\pi \left( \int_0^3 t \sqrt{\frac{t^2}{4} + 1} \, dt + \int_0^3 3 \sqrt{\frac{t^2}{4} + 1} \, dt \right).
   \]

4. **Evaluate the integrals:**

   Let's handle each part separately:

   **First integral:**
   \[
   I_1 = \int_0^3 t \sqrt{\frac{t^2}{4} + 1} \, dt.
   \]
   Use substitution:
   \[
   u = \frac{t^2}{4} + 1 \quad \Rightarrow \quad du = \frac{t}{2} \, dt \quad \Rightarrow \quad 2 \, du = t \, dt.
   \]
   Change the limits accordingly:
   When \( t = 0 \), \( u = 1 \).
   When \( t = 3 \), \( u = \frac{9}{4} + 1 = \frac{13}{4} \).

   Therefore,
   \[
   I_1 = \int_1^{\frac{13}{4}} 2u^{1/2} \, du = 2 \int_1^{\frac{13}{4}} u^{1/2} \, du.
   \]
   This evaluates to:
   \[
   2 \left[ \frac{2}{3} u^{3/2} \right]_1^{\frac{13}{4}} = \frac{4}{3} \left[ u^{3/2} \right]_1^{\frac{13}{4}} = \frac{4}{3} \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right).
   \]

   **Second integral:**
   \[
   I_2 = \int_0^3 3 \sqrt{\frac{t^2}{4} + 1} \, dt.
   \]
   Again, using the same substitution \( u = \frac{t^2}{4} + 1 \):
   \[
   I_2 = 3 \int_1^{\frac{13}{4}} 2u^{1/2} \, du = 6 \int_1^{\frac{13}{4}} u^{1/2} \, du = 6 \left[ \frac{2}{3} u^{3/2} \right]_1^{\frac{13}{4}} = 4 \left[ u^{3/2} \right]_1^{\frac{13}{4}} = 4 \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right).
   \]

5. **Combine the results:**

   \[
   2\pi \left( I_1 + I_2 \right) = 2\pi \left( \frac{4}{3} \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right) + 4 \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right) \right).
   \]

   Simplify the expression:
   \[
   2\pi \left( \frac{4}{3} + 4 \right) \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right) = 2\pi \left( \frac{4}{3} + \frac{12}{3} \right) \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right) = 2\pi \left( \frac{16}{3} \right) \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right).
   \]

   Therefore,
   \[
   \boxed{\frac{32\pi}{3} \left( \left( \frac{13}{4} \right)^{3/2} - 1 \right)}.
   \]

Would you like further details or have any questions?

Here are some more problems you can ask about:
1. Evaluate \(\int_0^2 x^2 e^x \, dx\).
2. Find the derivative of \(f(x) = x^3 \sin(x)\).
3. Solve the differential equation \(\frac{dy}{dx} = y \cos(x)\).
4. Calculate the limit \(\lim_{x \to 0} \frac{\sin(x)}{x}\).
5. Determine the convergence of the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\).

**Tip:** When solving integrals, substituting complex expressions can often simplify the problem, especially if it allows you to convert the integral into a standard form.

Learn how to solve the integral 2π∫₀³(t+3)√((t/2)²+1)dt with a detailed step-by-step solution. This problem involves calculus and integration techniques, including substitution methods. Suitable for college-level students and enthusiasts.

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