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Sia

Question
Math
Posted 10 months ago

Evaluate the iterated integral.
11(22ex+eydy)dx=
\int_{-1}^{1}\left(\int_{-2}^{2} e^{x}+e^{y} d y\right) d x=


Choose 1 answer:
(A) 2e2+4e4e12e22 e^{2}+4 e-4 e^{-1}-2 e^{-2}
(B) e24e4e1+e2e^{2}-4 e-4 e^{-1}+e^{-2}
(C) 4e2+4e24 e^{2}+4 e^{-2}
(D) 2e22e22 e^{2}-2 e^{-2}
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Answer from Sia
Posted 10 months ago
Solution by Steps
step 2
Since exe^{x} is a constant with respect to yy, the integral becomes ex22dy+22eydye^{x} \int_{-2}^{2} dy + \int_{-2}^{2} e^{y} \, dy
step 3
Compute the integrals: ex(y22)+(ey22)e^{x} \cdot (y \big|_{-2}^{2}) + (e^{y} \big|_{-2}^{2})
step 4
Substitute the limits of integration: ex(2(2))+(e2e2)e^{x} \cdot (2 - (-2)) + (e^{2} - e^{-2})
step 5
Simplify the expression: 4ex+(e2e2)4e^{x} + (e^{2} - e^{-2})
step 6
Evaluate the outer integral with respect to xx: 11(4ex+e2e2)dx\int_{-1}^{1} (4e^{x} + e^{2} - e^{-2}) \, dx
step 7
Split the integral: 411exdx+11(e2e2)dx4 \int_{-1}^{1} e^{x} \, dx + \int_{-1}^{1} (e^{2} - e^{-2}) \, dx
step 8
Since e2e2e^{2} - e^{-2} is constant with respect to xx, the second integral becomes (e2e2)(x11)(e^{2} - e^{-2}) \cdot (x \big|_{-1}^{1})
step 9
Compute the integrals: 4(ex11)+(e2e2)(1(1))4(e^{x} \big|_{-1}^{1}) + (e^{2} - e^{-2}) \cdot (1 - (-1))
step 10
Substitute the limits of integration: 4(e1e1)+(e2e2)24(e^{1} - e^{-1}) + (e^{2} - e^{-2}) \cdot 2
step 11
Simplify the expression: 4(ee1)+2(e2e2)4(e - e^{-1}) + 2(e^{2} - e^{-2})
step 12
Factor out common terms: 4e4e1+2e22e24e - 4e^{-1} + 2e^{2} - 2e^{-2}
step 13
Combine like terms: 2e2+4e4e12e22e^{2} + 4e - 4e^{-1} - 2e^{-2}
A
Key Concept
Iterated Integration
Explanation
When evaluating an iterated integral, we integrate with respect to one variable at a time while treating the other variable as a constant. After integrating with respect to the inner variable, we substitute the limits of integration for that variable, and then integrate the resulting expression with respect to the outer variable.

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