Question

Evaluate the double integral. ∬9xdA,D={(x,y)∣0≤x≤π,0≤y≤sin(x)}

Answer 1

To evaluate the given double integral ∬9xdA over the region D={(x,y)∣0≤x≤π,0≤y≤sin(x)}, we can write the integral in terms of x and evaluate it using integration by parts.

Step-by-step explanation:

To evaluate the given double integral ∬9xdA over the region D={(x,y)∣0≤x≤π,0≤y≤sin(x)}, we can write the integral in terms of x as:

∫[0 to π] ∫[0 to sin(x)] 9x dy dx

Integrating with respect to y first, we get:

∫[0 to π] (9xsin(x)) dx

Now, we can evaluate this integral by applying integration by parts.

Let u = x and dv = 9sin(x) dx

Then, du = dx and v = -9cos(x)

Using the formula for integration by parts:

∫(uv) dx = uv - ∫(v du)

Applying this formula to our integral, we get:

= -9xcos(x) - ∫(-9cos(x) dx)

= -9xcos(x) + 9sin(x) + C

So, the value of the double integral is -9πcos(π) + 9sin(π) + 9cos(0) - 9sin(0) = 9