Use the property to estimate the best possible bounds of the integral

Use the property to estimate the best possible bounds of the integral: fI 6 sin4(x + Y) dA, T is the triangle enclosed by the lines y = 0, Y = 4X, and x = 1_ < fI 6 sin4(x + Y) dA <

The Correct Answer and Explanation is:

To estimate the best possible bounds for the double integral $$\iint_T 6 \sin^4(x + y)\, dA,$$ where TT is the triangular region enclosed by the lines y=0y = 0, y=4xy = 4x, and x=1x = 1, we apply a standard bounding technique.

Step 1: Analyze the Region

The triangle TT has vertices at:

• (0,0)(0, 0) from the intersection of y=0y = 0 and y=4xy = 4x,
• (1,0)(1, 0) from x=1x = 1 and y=0y = 0,
• (1,4)(1, 4) from x=1x = 1 and y=4xy = 4x.

This forms a right triangle with base along the x-axis from x=0x = 0 to x=1x = 1 and height rising up to y=4y = 4 at x=1x = 1.

The area of this triangle is:

Area=12×base×height=12×1×4=2.\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 4 = 2.

Step 2: Understand the Integrand

The integrand is 6sin⁡4(x+y)6 \sin^4(x + y). Since sine values always lie in [−1,1][-1, 1], raising sine to the fourth power produces values in [0,1][0, 1]. Multiplying by 6 scales this to the range [0,6][0, 6]. Therefore, across the entire region TT, the function 6sin⁡4(x+y)6 \sin^4(x + y) takes values between 0 and 6.

Step 3: Apply the Bounding Property

We now use the property that if m≤f(x,y)≤Mm \leq f(x, y) \leq M for all (x,y)∈T(x, y) \in T, then:

m⋅Area(T)≤∬Tf(x,y) dA≤M⋅Area(T).m \cdot \text{Area}(T) \leq \iint_T f(x, y)\, dA \leq M \cdot \text{Area}(T).

Substituting m=0m = 0, M=6M = 6, and Area=2\text{Area} = 2:

0≤∬T6sin⁡4(x+y) dA≤12.0 \leq \iint_T 6 \sin^4(x + y)\, dA \leq 12.

Final Answer

0<∬T6sin⁡4(x+y) dA<12\boxed{0 < \iint_T 6 \sin^4(x + y)\, dA < 12}

This result gives the best possible bounds without evaluating the integral directly.