Let's solve the provided problems step by step.

2. Evaluate the following integrals:

(a) $\int 15x \, dx$

Using the basic integration rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$

Here: $\int 15x dx = 15 \cdot \frac{x^2}{2} + C = \frac{15x^2}{2} + C$

(b) $\int (3x^2 - 4) \, dx$

Integrate term by term: $\int 3x^2 dx = \frac{3x^{3}}{3} = x^3$ $\int -4 dx = -4x$

So: $\int (3x^2 - 4) dx = x^3 - 4x + C$

(c) $\int \frac{3}{x^4} \, dx$

Rewrite $\frac{1}{x^4}$ as $x^{-4}$, then use the power rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$

Here: $\int 3x^{-4} dx = 3 \cdot \frac{x^{-3}}{-3} + C = -\frac{1}{x^3} + C$

(d) $\int (5x^2 + 2\cos(x)) \, dx$

Integrate term by term: $\int 5x^2 dx = \frac{5x^3}{3}$ $\int 2\cos(x) dx = 2\sin(x)$

So: $\int (5x^2 + 2\cos(x)) dx = \frac{5x^3}{3} + 2\sin(x) + C$

3. By using the substitution $t = 3x + 1$, find $\int 12(3x+1)^3 dx$

Let $t = 3x + 1$, then $dt = 3dx \rightarrow dx = \frac{dt}{3}$.

The integral becomes: $\int 12(3x+1)^3 dx = \int 12t^3 \cdot \frac{dt}{3} = 4 \int t^3 dt$

Now integrate $t^3$: $\int t^3 dt = \frac{t^4}{4}$

So: $4 \int t^3 dt = 4 \cdot \frac{t^4}{4} = t^4 + C$

Substitute back $t = 3x + 1$: $\int 12(3x+1)^3 dx = (3x + 1)^4 + C$

Would you like further explanation for any step, or help with problem 4?

Expanding Understanding

Here are 5 related questions to deepen your understanding:

1. How do we apply substitution when integrating more complex expressions?
2. Can you derive the general power rule for integration from scratch?
3. What are some practical uses of definite integrals in real-world applications?
4. How can you confirm your integration result is correct by differentiating it?
5. How do you handle integrals with multiple substitutions or complex trigonometric identities?

Tip

Always double-check integration results by differentiating your answer; this ensures that it matches the original integrand!