58.0k views
5 votes
Evaluate the iterated integral ∫[3 to 5] ∫[0 to π/2] x * sin(y) dy dx.

User Van Peer
by
8.4k points

1 Answer

1 vote

Final answer:

To evaluate the iterated integral ∫[3 to 5] ∫[0 to π/2] x * sin(y) dy dx, we can evaluate the inner integral with respect to y first, and then the outer integral with respect to x. The value of the iterated integral is 8.

Step-by-step explanation:

To evaluate the iterated integral ∫[3 to 5] ∫[0 to π/2] x * sin(y) dy dx, we can use the properties of iterated integrals. First, we integrate the inner integral with respect to y. ∫[0 to π/2] x * sin(y) dy = -x * cos(y) evaluated from y = 0 to y = π/2 = -x * (cos(π/2) - cos(0)) = -x * (0 - 1) = x

Now, we integrate the outer integral with respect to x. ∫[3 to 5] x dx = (1/2) * x^2 evaluated from x = 3 to x = 5 = (1/2) * (5^2 - 3^2) = (1/2) * (25 - 9) = 8

Therefore, the value of the iterated integral is 8.

User Antony Koch
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories