If ∫−1−4f(x)dx=0 and ∫31g(x)dx=3, what is the value of ∫∫df(x)g(y)da where d is the square: −4≤x≤−1, 1≤y≤3?

0

Step-by-step explanation:

From your problem, we have to extract the information that are important from the first two intregrals so we can solve the double integral.

We also have that:

---------------------------------

With this, now we can solve the double integral.

Since the limits of integration are constant, i can use dA both as dydx or dxdy. I am going to use dydx.

So the double integral will be:

We solve a double integral from the inside to the outside, so the first integral we solve is:

f is a function of x and we are integrating dy, so this means that f is a constant. Our integral now is this:

From above, we have that

So,

Now we have to solve the outside integral:

We know that

So the double integral will be 0

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: