"Evaluate the following iterated integrals. ∫03∫−21(2x+3y)dxdy" was answered by our community

djeljenike

Answered question

2021-03-02

Evaluate the following iterated integrals.

\int_{0}^{3} \int_{- 2}^{1} (2 x + 3 y) d x d y

comentezq

Skilled2021-03-03Added 106 answers

Given integral,

\int_{0}^{3} \int_{- 2}^{1} (2 x + 3 y) d x d y

Now

\int_{0}^{3} \int_{- 2}^{1} (2 x + 3 y) d x d y = \int_{0}^{3} (\frac{2 x^{2}}{2} + 3 x y)_{- 2}^{1} d y

= \int_{0}^{3} (x^{2} + 3 x y)_{- 2}^{1} d y

= \int_{0}^{3} ((1)^{2} + 3 (1) y - (- 2)^{2} - 3 (- 2) y) d y

= \int_{0}^{3} (1 + 3 y - 4 + 6 y) d y

= \int_{0}^{3} (- 3 + 9 y) d y

= (- 3 y + \frac{9 y^{2}}{2})_{0}^{3}

= (- 3 (3) + \frac{9 (3)^{2}}{2} + 3 (0) + \frac{9 (0)^{2}}{2})

= - 9 + \frac{81}{2} + 0

= \frac{63}{2}

Final answer:

\int_{0}^{3} \int_{- 2}^{1} (2 x + 3 y) d x d y = \frac{63}{2}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?