Question

integrate \\(\int_{-2}^{2} 21(3x + 2)^{6} dx\\)
answer =

Explanation:

Step1: Use substitution method

Let $u = 3x + 2$. Then $du = 3dx$, so $dx = \frac{du}{3}$.
Adjust bounds: When $x=-2$, $u=3(-2)+2=-4$; when $x=2$, $u=3(2)+2=8$.
Rewrite integral:
$$\int_{-4}^{8} 21u^6 \cdot \frac{du}{3} = \int_{-4}^{8} 7u^6 du$$

Step2: Apply power rule for integration

Power rule: $\int u^n du = \frac{u^{n+1}}{n+1} + C$
$$7 \cdot \frac{u^{7}}{7} \bigg|_{-4}^{8} = u^7 \bigg|_{-4}^{8}$$

Step3: Evaluate at bounds

Substitute upper and lower limits:
$$8^7 - (-4)^7$$
Calculate values: $8^7=2097152$, $(-4)^7=-16384$
$$2097152 - (-16384) = 2097152 + 16384$$

Answer:

2113536