Use the property of integrals to estimate the best possible bounds of the integral ∫∫τ 9 sin(x + y) dA, where T is a triangle enclosed by y = 0, y = 5x and x = 4.

Solution:

The concepts of lower bounds and upper bounds in integration are very interesting as well as tricky. The lower bound represents the smallest value from which we start summing areas (smallest value of the interval) and the upper bound is the value to which we sum to (maximum value of the interval). We will look at an example and use the property of bounds indefinite integrals to solve the question.

Let's have a look at the explanation to understand better. To solve this question, we use the property of bounds.

It says that: 

⇒ If m ≤ f(x, y) ≤ M for all (x, y) in D, then, m × A(D) ≤ ∫∫τ f(x, y) dA ≤ M × A(D)

Here, we have f(x, y) = 9 sin(x + y).

⇒ We know that minimum and maximum values of sine are -1 and 1 respectively.

⇒ Hence, we have -9 ≤ 9 sin(x + y) ≤ 9.

⇒ Now, we find the area A(D) enclosed by the triangle bounded by the lines given, which is 1/2 × 20 × 4 = 40.

⇒ Now, we use the property of bounds to get, -9 × 40 ≤ ∫∫τ 9 sin(x + y) dA ≤ 9 × 40 = -360 ≤ ∫∫τ 9 sin(x + y) dA ≤ 360.

Hence, the lower bound to the integral ∫∫τ 9 sin(x + y) dA is -360, and the upper bound for the same is 360.

Use the property of integrals to estimate the best possible bounds of the integral ∫∫τ 9 sin(x + y) dA, where T is a triangle enclosed by y = 0, y = 5x and x = 4.

Summary:

The lower bound to the integral ∫∫τ 9 sin(x + y) dA is -360, and the upper bound for the same is 360.