$$ \int_{1}^{4} \int_{0}^{\pi} x \sin (y) d y d x=A $$ Then the value of $$ A $$ is

Question

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To evaluate the double integral

$$
A = \int_{1}^{4} \int_{0}^{\pi} x \sin(y) \, dy \, dx,
$$

we can first compute the inner integral with respect to $$ y $$:

$$
\int_{0}^{\pi} x \sin(y) \, dy.
$$

Since $$ x $$ is treated as a constant with respect to $$ y $$, we can factor it out of the integral:

$$
\int_{0}^{\pi} x \sin(y) \, dy = x \int_{0}^{\pi} \sin(y) \, dy.
$$

Next, we need to evaluate the integral $$ \int_{0}^{\pi} \sin(y) \, dy $$:

$$
\int_{0}^{\pi} \sin(y) \, dy = -\cos(y) \bigg|_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2.
$$

Thus, we have:

$$
\int_{0}^{\pi} x \sin(y) \, dy = x \cdot 2 = 2x.
$$

Now we substitute this result back into the outer integral:

$$
A = \int_{1}^{4} 2x \, dx.
$$

Next, we compute the integral $$ \int_{1}^{4} 2x \, dx $$:

$$
\int 2x \, dx = x^2 + C.
$$

Now we evaluate it from 1 to 4:

$$
\int_{1}^{4} 2x \, dx = \left[ x^2 \right]_{1}^{4} = 4^2 - 1^2 = 16 - 1 = 15.
$$

Thus, the value of $$ A $$ is

$$
\boxed{15}.
$$
The value of $$ A $$ is 15.

Then the value of is