Answer :
Of course! To solve the double integral
[tex]\[ \int_1^4 \int_1^8 \left(3 x^2 y - x y\right) \, dy \, dx, \][/tex]
let’s break it down into steps, performing the inner integral first with respect to [tex]\(y\)[/tex], and then the outer integral with respect to [tex]\(x\)[/tex].
1. Inner Integral: Integrate the function [tex]\(3 x^2 y - x y\)[/tex] with respect to [tex]\(y\)[/tex] from [tex]\(1\)[/tex] to [tex]\(8\)[/tex]:
[tex]\[ \int_1^8 (3 x^2 y - x y) \, dy. \][/tex]
To do this, integrate each term separately:
- First term: [tex]\(\int_1^8 3 x^2 y \, dy\)[/tex]:
[tex]\[ 3 x^2 \int_1^8 y \, dy = 3 x^2 \left[ \frac{y^2}{2} \right]_1^8 = 3 x^2 \left( \frac{8^2}{2} - \frac{1^2}{2} \right) = 3 x^2 \left( \frac{64}{2} - \frac{1}{2} \right) = 3 x^2 \left( 32 - \frac{1}{2} \right) = 3 x^2 \frac{63}{2} = \frac{189 x^2}{2}. \][/tex]
- Second term: [tex]\(\int_1^8 x y \, dy\)[/tex]:
[tex]\[ x \int_1^8 y \, dy = x \left[ \frac{y^2}{2} \right]_1^8 = x \left( \frac{8^2}{2} - \frac{1^2}{2} \right) = x \left( \frac{64}{2} - \frac{1}{2} \right) = x \left( 32 - \frac{1}{2} \right) = x \frac{63}{2} = \frac{63 x}{2}. \][/tex]
Combining these results, the inner integral becomes:
[tex]\[ \int_1^8 (3 x^2 y - x y) \, dy = \frac{189 x^2}{2} - \frac{63 x}{2} = \frac{189 x^2}{2} - \frac{63 x}{2}. \][/tex]
2. Outer Integral: Now, integrate the result of the inner integral with respect to [tex]\(x\)[/tex] from [tex]\(1\)[/tex] to [tex]\(4\)[/tex]:
[tex]\[ \int_1^4 \left( \frac{189 x^2}{2} - \frac{63 x}{2} \right) \, dx. \][/tex]
Again, integrate each term separately:
- First term: [tex]\(\int_1^4 \frac{189 x^2}{2} \, dx\)[/tex]:
[tex]\[ \frac{189}{2} \int_1^4 x^2 \, dx = \frac{189}{2} \left[ \frac{x^3}{3} \right]_1^4 = \frac{189}{2} \left( \frac{4^3}{3} - \frac{1^3}{3} \right) = \frac{189}{2} \left( \frac{64}{3} - \frac{1}{3} \right) = \frac{189}{2} \left( \frac{63}{3} \right) = \frac{189}{2} \times 21 = 189 \times 10.5 = 1984.5. \][/tex]
- Second term: [tex]\(\int_1^4 \frac{63 x}{2} \, dx\)[/tex]:
[tex]\[ \frac{63}{2} \int_1^4 x \, dx = \frac{63}{2} \left[ \frac{x^2}{2} \right]_1^4 = \frac{63}{2} \left( \frac{4^2}{2} - \frac{1^2}{2} \right) = \frac{63}{2} \left( \frac{16}{2} - \frac{1}{2} \right) = \frac{63}{2} \left( 8 - \frac{1}{2} \right) = \frac{63}{2} \left( \frac{15}{2} \right) = \frac{63}{2} \times 7.5 = 472.5. \][/tex]
Finally, combining the results of the outer integral:
[tex]\[ \int_1^4 \left( \frac{189 x^2}{2} - \frac{63 x}{2} \right) \, dx = 1984.5 - 472.5 = 1512. \][/tex]
The value of the double integral is:
[tex]\[ \int_1^4 \int_1^8 \left( 3 x^2 y - x y \right) \, dy \, dx = 1748.25, \text{ which simplifies to } \frac{6993}{4}. \][/tex]
So, the final answer is:
[tex]\[ \boxed{\frac{6993}{4}}. \][/tex]
[tex]\[ \int_1^4 \int_1^8 \left(3 x^2 y - x y\right) \, dy \, dx, \][/tex]
let’s break it down into steps, performing the inner integral first with respect to [tex]\(y\)[/tex], and then the outer integral with respect to [tex]\(x\)[/tex].
1. Inner Integral: Integrate the function [tex]\(3 x^2 y - x y\)[/tex] with respect to [tex]\(y\)[/tex] from [tex]\(1\)[/tex] to [tex]\(8\)[/tex]:
[tex]\[ \int_1^8 (3 x^2 y - x y) \, dy. \][/tex]
To do this, integrate each term separately:
- First term: [tex]\(\int_1^8 3 x^2 y \, dy\)[/tex]:
[tex]\[ 3 x^2 \int_1^8 y \, dy = 3 x^2 \left[ \frac{y^2}{2} \right]_1^8 = 3 x^2 \left( \frac{8^2}{2} - \frac{1^2}{2} \right) = 3 x^2 \left( \frac{64}{2} - \frac{1}{2} \right) = 3 x^2 \left( 32 - \frac{1}{2} \right) = 3 x^2 \frac{63}{2} = \frac{189 x^2}{2}. \][/tex]
- Second term: [tex]\(\int_1^8 x y \, dy\)[/tex]:
[tex]\[ x \int_1^8 y \, dy = x \left[ \frac{y^2}{2} \right]_1^8 = x \left( \frac{8^2}{2} - \frac{1^2}{2} \right) = x \left( \frac{64}{2} - \frac{1}{2} \right) = x \left( 32 - \frac{1}{2} \right) = x \frac{63}{2} = \frac{63 x}{2}. \][/tex]
Combining these results, the inner integral becomes:
[tex]\[ \int_1^8 (3 x^2 y - x y) \, dy = \frac{189 x^2}{2} - \frac{63 x}{2} = \frac{189 x^2}{2} - \frac{63 x}{2}. \][/tex]
2. Outer Integral: Now, integrate the result of the inner integral with respect to [tex]\(x\)[/tex] from [tex]\(1\)[/tex] to [tex]\(4\)[/tex]:
[tex]\[ \int_1^4 \left( \frac{189 x^2}{2} - \frac{63 x}{2} \right) \, dx. \][/tex]
Again, integrate each term separately:
- First term: [tex]\(\int_1^4 \frac{189 x^2}{2} \, dx\)[/tex]:
[tex]\[ \frac{189}{2} \int_1^4 x^2 \, dx = \frac{189}{2} \left[ \frac{x^3}{3} \right]_1^4 = \frac{189}{2} \left( \frac{4^3}{3} - \frac{1^3}{3} \right) = \frac{189}{2} \left( \frac{64}{3} - \frac{1}{3} \right) = \frac{189}{2} \left( \frac{63}{3} \right) = \frac{189}{2} \times 21 = 189 \times 10.5 = 1984.5. \][/tex]
- Second term: [tex]\(\int_1^4 \frac{63 x}{2} \, dx\)[/tex]:
[tex]\[ \frac{63}{2} \int_1^4 x \, dx = \frac{63}{2} \left[ \frac{x^2}{2} \right]_1^4 = \frac{63}{2} \left( \frac{4^2}{2} - \frac{1^2}{2} \right) = \frac{63}{2} \left( \frac{16}{2} - \frac{1}{2} \right) = \frac{63}{2} \left( 8 - \frac{1}{2} \right) = \frac{63}{2} \left( \frac{15}{2} \right) = \frac{63}{2} \times 7.5 = 472.5. \][/tex]
Finally, combining the results of the outer integral:
[tex]\[ \int_1^4 \left( \frac{189 x^2}{2} - \frac{63 x}{2} \right) \, dx = 1984.5 - 472.5 = 1512. \][/tex]
The value of the double integral is:
[tex]\[ \int_1^4 \int_1^8 \left( 3 x^2 y - x y \right) \, dy \, dx = 1748.25, \text{ which simplifies to } \frac{6993}{4}. \][/tex]
So, the final answer is:
[tex]\[ \boxed{\frac{6993}{4}}. \][/tex]
