Question

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?

Answer 1

Answer:

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.

$\int\limits^{-1}_{-4} {f(x)} \, dx = 0$

We also have that:

$\int\limits^{3}_{1} {g(x)} \, dx = 3$

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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:

$\int \limits^{-1}_{-4} \int \limits^{3}_{1} {g(y)} {f(x)} dy dx$

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

$\int \limits^{3}_{1} \y g(y) \x f(x) dy$

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

$\x f(x) \int \limits^{3}_{1} \y g(y) dy$

From above, we have that

$\int \limits^{3}_{1} \x g(x) dx = 3$

So,

$\int \limits^{3}_{1} \y g(y) dy = 3$

Now we have to solve the outside integral:

$3\int\limits^{-1}_{-4} {f(x)} \, dx$

We know that

$\int\limits^{-1}_{-4} {f(x)} \, dx = 0$

So the double integral will be 0