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Calculus IIIChapter 4: Multiple IntegralsSection 4-1

Double Integrals — Question 5

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Question 5

Let R=[−1,2]×[0,2]R=[-1,2]\times[0,2]. A continuous function ff satisfies 2≤f(x,y)≤5,∬RfdA=19.2\le f(x,y)\le 5,\qquad \iint_R f\,dA=19. A second integrable function gg satisfies |g(x,y)−f(x,y)|≤0.1|g(x,y)-f(x,y)|\le 0.1 throughout RR.

Tasks

  1. Check that the stated integral of ff is consistent with its pointwise bounds.

  2. Evaluate ∬R(3f−4)dA\iint_R(3f-4)\,dA.

  3. Give the sharp bounds guaranteed by the data for ∬RgdA\iint_Rg\,dA.

Original worksheet page 1: question and worked solution for 4-1-005
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Question 5 – Solution

Strategy. Convert pointwise inequalities to integral inequalities using the area of the rectangle, and use linearity for the transformed function.

Step 1: Comparison bounds The rectangle has area A=(2−(−1))(2−0)=6.A=(2-(-1))(2-0)=6. Integrating 2≤f≤52\le f\le 5 over RR gives 12≤∬RfdA≤30.12\le\iint_R f\,dA\le 30. The given value 1919 lies in this interval, so the data are consistent.

Step 2: Linearity Because the integral of the constant 44 is 4A=244A=24, ∬R(3f−4)dA=3∬RfdA−∬R4dA=57−24=33.\iint_R(3f-4)\,dA =3\iint_R f\,dA-\iint_R4\,dA =57-24=\boxed{33}.

Step 3: Stability bound From |g−f|≤0.1|g-f|\le 0.1, −0.1≤g−f≤0.1.-0.1\le g-f\le 0.1. Integration over an area-66 region yields −0.6≤∬R(g−f)dA≤0.6.-0.6\le\iint_R(g-f)\,dA\le 0.6. Consequently, 18.4≤∬RgdA≤19.6.\boxed{18.4\le\iint_Rg\,dA\le 19.6}. These endpoints are attainable by g=f−0.1g=f-0.1 and g=f+0.1g=f+0.1, so no tighter bounds follow from the supplied information.

Original worksheet page 2: question and worked solution for 4-1-005

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