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.