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Evaluate the double integral:

[tex]\[ \int_1^4 \int_1^8 \left(3 x^2 y - xy\right) \, dy \, dx \][/tex]

Sagot :

Sure, let's solve the integral [tex]\(\int_1^4 \int_1^8 (3 x^2 y - x y) \, dy \, dx\)[/tex] step by step.

### Step 1: Integrate the inner integral with respect to [tex]\(y\)[/tex]

First, consider the inner integral:

[tex]\[ \int_1^8 (3 x^2 y - x y) \, dy \][/tex]

To do this, we integrate each term separately with respect to [tex]\(y\)[/tex].

[tex]\[ \int_1^8 3 x^2 y \, dy = 3 x^2 \int_1^8 y \, dy \][/tex]

The integral of [tex]\(y\)[/tex] with respect to [tex]\(y\)[/tex] is:

[tex]\[ \int y \, dy = \frac{y^2}{2} \][/tex]

Evaluating it from [tex]\(1\)[/tex] to [tex]\(8\)[/tex]:

[tex]\[ \left[ \frac{y^2}{2} \right]_1^8 = \frac{8^2}{2} - \frac{1^2}{2} = \frac{64}{2} - \frac{1}{2} = 32 - 0.5 = 31.5 \][/tex]

Hence,

[tex]\[ 3 x^2 \int_1^8 y \, dy = 3 x^2 \cdot 31.5 = 94.5 x^2 \][/tex]

Similarly, integrate [tex]\(x y\)[/tex] with respect to [tex]\(y\)[/tex]:

[tex]\[ \int_1^8 x y \, dy = x \int_1^8 y \, dy = x \cdot 31.5 = 31.5 x \][/tex]

### Putting it all together:

[tex]\[ \int_1^8 (3 x^2 y - x y) \, dy = 94.5 x^2 - 31.5 x \][/tex]

### Step 2: Integrate the result with respect to [tex]\(x\)[/tex]

Now we consider the outer integral:

[tex]\[ \int_1^4 (94.5 x^2 - 31.5 x) \, dx \][/tex]

We integrate each term separately with respect to [tex]\(x\)[/tex]:

[tex]\[ \int_1^4 94.5 x^2 \, dx \][/tex]

The integral of [tex]\(x^2\)[/tex] with respect to [tex]\(x\)[/tex] is:

[tex]\[ \int x^2 \, dx = \frac{x^3}{3} \][/tex]

So we have:

[tex]\[ 94.5 \int x^2 \, dx = 94.5 \left[ \frac{x^3}{3} \right]_1^4 = 94.5 \cdot \frac{1}{3} \left[ 64 - 1 \right] = 31.5 \left[ 64 - 1 \right] = 31.5 \cdot 63 = 1984.5 \][/tex]

Next, we integrate [tex]\(31.5 x\)[/tex]:

[tex]\[ \int_1^4 31.5 x \, dx \][/tex]

The integral of [tex]\(x\)[/tex] with respect to [tex]\(x\)[/tex] is:

[tex]\[ \int x \, dx = \frac{x^2}{2} \][/tex]

So we have:

[tex]\[ 31.5 \int x \, dx = 31.5 \left[ \frac{x^2}{2} \right]_1^4 = 31.5 \cdot \frac{1}{2} \left[ 16 - 1 \right] = 15.75 \left[ 16 - 1 \right] = 15.75 \cdot 15 = 236.25 \][/tex]

### Putting everything together:

[tex]\[ \int_1^4 (94.5 x^2 - 31.5 x) \, dx = 1984.5 - 236.25 = 1748.25 \][/tex]

### Final Answer:

[tex]\[ \boxed{\frac{6993}{4}} \][/tex]

Which simplifies to [tex]\(1748.25\)[/tex].