Evaluate the double integral ${\int }_0^6{\int }_0^1(6x+1+xy) dy dx$.

• A. $246$
• B. $234$
• C. $258$
• D. $222$

Answer: (A)

Verified step by step guidance

1. Set up the double integral as \( \int_0^6 \int_0^1 (6x + 1 + xy) \, dy \, dx \). The outer integral is with respect to \(x\) from 0 to 6, and the inner integral is with respect to \(y\) from 0 to 1.
2. First, focus on the inner integral \( \int_0^1 (6x + 1 + xy) \, dy \). Break it into separate terms: \( \int_0^1 6x \, dy + \int_0^1 1 \, dy + \int_0^1 xy \, dy \).
3. Evaluate each term of the inner integral: For \( \int_0^1 6x \, dy \), treat \(6x\) as a constant with respect to \(y\), so the result is \(6x \cdot y \big|_0^1 = 6x(1) - 6x(0) = 6x\). For \( \int_0^1 1 \, dy \), the result is \(y \big|_0^1 = 1 - 0 = 1\). For \( \int_0^1 xy \, dy \), treat \(x\) as a constant, so the result is \(x \cdot \frac{y^2}{2} \big|_0^1 = x \cdot \frac{1^2}{2} - x \cdot \frac{0^2}{2} = \frac{x}{2}\).
4. Combine the results of the inner integral: \(6x + 1 + \frac{x}{2}\). This simplifies to \(6x + \frac{x}{2} + 1 = \frac{12x}{2} + \frac{x}{2} + 1 = \frac{13x}{2} + 1\).
5. Now, evaluate the outer integral \( \int_0^6 \left( \frac{13x}{2} + 1 \right) \, dx \). Break it into two terms: \( \int_0^6 \frac{13x}{2} \, dx + \int_0^6 1 \, dx \). For \( \int_0^6 \frac{13x}{2} \, dx \), the result is \( \frac{13}{2} \cdot \frac{x^2}{2} \big|_0^6 \). For \( \int_0^6 1 \, dx \), the result is \(x \big|_0^6 = 6 - 0 = 6\). Combine these results to find the final value of the double integral.