Calculate the value of the double integral ${\int }_0^4{\int }_0^1(4x+1+xy) dy dx$.

• A. $44$
• B. $36$
• C. $40$
• D. $48$

Answer: (A)

Verified step by step guidance

1. Step 1: Understand the problem. The given double integral is \( \int_0^4 \int_0^1 (4x + 1 + xy)\, dy\, dx \). This means we are integrating the function \( 4x + 1 + xy \) first with respect to \( y \) (inner integral) and then with respect to \( x \) (outer integral).
2. Step 2: Perform the inner integral. To integrate \( 4x + 1 + xy \) with respect to \( y \), treat \( x \) as a constant. The integral becomes \( \int_0^1 (4x + 1 + xy)\, dy \). Break it into separate terms: \( \int_0^1 4x\, dy + \int_0^1 1\, dy + \int_0^1 xy\, dy \).
3. Step 3: Compute each term of the inner integral. For \( \int_0^1 4x\, dy \), the result is \( 4x \cdot y \) evaluated from \( y = 0 \) to \( y = 1 \). For \( \int_0^1 1\, dy \), the result is \( y \) evaluated from \( y = 0 \) to \( y = 1 \). For \( \int_0^1 xy\, dy \), the result is \( x \cdot \frac{y^2}{2} \) evaluated from \( y = 0 \) to \( y = 1 \).
4. Step 4: Combine the results of the inner integral. Substitute the limits of \( y \) into each term: \( 4x \cdot (1 - 0) + (1 - 0) + x \cdot \frac{1^2}{2} \). Simplify this expression to get \( 4x + 1 + \frac{x}{2} \). This is the result of the inner integral.
5. Step 5: Perform the outer integral. Now integrate \( \int_0^4 (4x + 1 + \frac{x}{2})\, dx \). Break it into separate terms: \( \int_0^4 4x\, dx + \int_0^4 1\, dx + \int_0^4 \frac{x}{2}\, dx \). Compute each term by applying the power rule for integration and evaluate the limits from \( x = 0 \) to \( x = 4 \). Combine the results to find the final value of the double integral.