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

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

the correct answer to your question is a. 4^4

good luck my friend!

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