Let's solve questions 11 and 13 step by step.

Question 11:

$\int 3x^2 e^{2x^3} \, dx$

Step 1: Substitution

Notice that the exponent $2x^3$ suggests using the substitution method. Let's set: $u = 2x^3 \quad \Rightarrow \quad du = 6x^2 \, dx \quad \Rightarrow \quad \frac{du}{6} = x^2 \, dx$

Now, rewrite the integral: $\int 3x^2 e^{2x^3} \, dx = \int 3 \cdot \frac{du}{6} e^u = \frac{1}{2} \int e^u \, du$

Step 2: Integrating

The integral of $e^u$ is straightforward: $\frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C$

Step 3: Back-Substitute

Now, substitute $u = 2x^3$ back into the solution: $\frac{1}{2} e^u + C = \frac{1}{2} e^{2x^3} + C$

Thus, the solution to the integral is: $\boxed{\frac{1}{2} e^{2x^3} + C}$

Question 13:

$\int (1 - t) e^{2t - t^2} \, dt$

Step 1: Substitution

Here, the exponent $2t - t^2$ suggests using substitution. Set: $u = 2t - t^2 \quad \Rightarrow \quad du = (2 - 2t) \, dt = -2(t - 1) \, dt \quad \Rightarrow \quad \frac{-du}{2} = (t - 1) \, dt$

Now, rewrite the integral: $\int (1 - t) e^{2t - t^2} \, dt = -\frac{1}{2} \int e^u \, du$

Step 2: Integrating

The integral of $e^u$ is again straightforward: $-\frac{1}{2} \int e^u \, du = -\frac{1}{2} e^u + C$

Step 3: Back-Substitute

Substitute $u = 2t - t^2$ back into the solution: $-\frac{1}{2} e^u + C = -\frac{1}{2} e^{2t - t^2} + C$

Thus, the solution to the integral is: $\boxed{-\frac{1}{2} e^{2t - t^2} + C}$

Summary of Solutions:

11. $\boxed{\frac{1}{2} e^{2x^3} + C}$
12. $\boxed{-\frac{1}{2} e^{2t - t^2} + C}$

Would you like further explanations or details on any steps?

Follow-up Questions:

1. How do you identify when substitution is a good method to use?
2. Why is the derivative of $x^3$ important in question 11?
3. How would these integrals change if the exponents were more complex, such as higher-order polynomials?
4. Can you generalize the steps for solving integrals involving $e^{f(x)}$?
5. What if the constant in question 11 was different, like $e^{3x^3}$? How would that change the approach?

Tip:

When dealing with exponentials like $e^{f(x)}$, look for a substitution that simplifies $f(x)$ to a linear form for easier integration!