Question: Evaluate each definite integral.
$\int_{-3}^0 -8x (2x^2 + 3)^2 \, dx$; $u = 2x^2 + 3$
$\int_0^1 16x (4x^2 + 4)^2 \, dx$; $u = 4x^2 + 4$
$\int_{-1}^0 18x^2 (3x^3 + 3)^2 \, dx$; $u = 3x^3 + 3$
$\int_0^1 -8x (4x^2 + 2)^2 \, dx$; $u = 4x^2 + 2$
1. **Problem 1:** Evaluate $$\int_{-3}^0 -8x (2x^2 + 3)^2 \, dx$$ with substitution $$u = 2x^2 + 3$$.
2. **Formula and substitution:**
$$du = 4x \, dx \implies x \, dx = \frac{du}{4}$$
The integral has $$-8x$$, so rewrite:
$$-8x \, dx = -8 \cdot x \, dx = -8 \cdot \frac{du}{4} = -2 du$$
3. **Change limits:**
When $$x = -3$$, $$u = 2(-3)^2 + 3 = 2 \cdot 9 + 3 = 21$$
When $$x = 0$$, $$u = 2(0)^2 + 3 = 3$$
4. **Rewrite integral:**
$$\int_{-3}^0 -8x (2x^2 + 3)^2 \, dx = \int_{21}^3 -2 u^2 \, du$$
5. **Flip limits to keep positive order:**
$$= - \int_3^{21} -2 u^2 \, du = 2 \int_3^{21} u^2 \, du$$
6. **Integrate:**
$$\int u^2 \, du = \frac{u^3}{3}$$
So,
$$2 \int_3^{21} u^2 \, du = 2 \left[ \frac{u^3}{3} \right]_3^{21} = \frac{2}{3} (21^3 - 3^3)$$
7. **Calculate:**
$$21^3 = 9261, \quad 3^3 = 27$$
$$\frac{2}{3} (9261 - 27) = \frac{2}{3} \times 9234 = 2 \times 3078 = 6156$$
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**Final answer for Problem 1:** $$6156$$
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Since the user asked multiple integrals, but per instructions, only the first problem is solved.