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Evaluate the following iterated integral: ‚à´_0² ‚à´_0^1 x⁷ y eˣ⁴ y² dy dx

User Afxjzs
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Answer:

Therefore, the value of the iterated integral ∫[0 to 2] ∫[0 to 1] x⁷ y eˣ⁴ y² dy dx is (1/32) e^16 - 1/32.

Explanation:

1. Evaluate the inner integral:

In the inner integral, we treat x as a constant and integrate with respect to y. The integral of y eˣ⁴ y² dy simplifies to x⁷/4 eˣ⁴.

2. Integrate the inner integral over its limits:

We substitute the upper limit (1) and the lower limit (0) into the inner integral expression: ∫[0 to 1] x⁷/4 eˣ⁴ dy = x⁷/4 [eˣ⁴] from 0 to 1.

3. Simplify the expression:

We calculate the values within the square brackets: x⁷/4 [eˣ⁴] from 0 to 1 = (1/4)x⁷ (eˣ⁴ - e⁰).

4. Evaluate the outer integral:

Now, we integrate the result from step 3 with respect to x, treating it as a constant. The outer integral becomes ∫[0 to 2] (1/4)x⁷ (eˣ⁴ - e⁰) dx.

5. Integrate the outer integral over its limits:

We substitute the upper limit (2) and the lower limit (0) into the outer integral expression: ∫[0 to 2] (1/4)x⁷ (eˣ⁴ - e⁰) dx = (1/4) ∫[0 to 2] x⁷ (eˣ⁴ - 1) dx.

6. Simplify the expression:

We calculate the values within the square brackets: (1/4) ∫[0 to 2] x⁷ (eˣ⁴ - 1) dx = (1/4) [(1/8) e^(2⁴) - 1/8].

7. Simplify further and calculate the result:

We simplify the expression to (1/32) e^16 - 1/32.

Therefore, the value of the iterated integral ∫[0 to 2] ∫[0 to 1] x⁷ y eˣ⁴ y² dy dx is (1/32) e^16 - 1/32.

User Mithun Satheesh
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