(a) Explanation Given information: The graph of f ( x ) is above the x - axis and concave down on the interval a 0 ≤ x ≤ a 1 . Let x 1 be the midpoint of this interval and Δ x = a 1 − a 0 Formula used: If base length and angle between the base and hypotenuse of a right angle triangle is same, then the area of the two right angled triangles are equal. Proof: There is a tangent line at ( x 1 , f ( x 1 ) ) to the function f ( x ) which forms two right triangles as shown in figure (b). Call the above triangle as T 1 and the below triangle as T 2 The base length of both the triangles T 1 and T 2 is same, that is Δ x 2 and angle between the base and hypotenuse of triangles is same (b) To determine Why T ≤ ∫ a b f ( x ) d x ≤ M , where T and M are the approximations given by the trapezoid and the midpoint rules respectively.

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Chapter 9.4, Problem 33E

(a)

To determine

To prove: The area of shaded trapezoid in figure (a) is the same as the area of the shaded rectangle in figure (c), that is f(x1)Δx

(b)

To determine

Why T≤∫abf(x)dx≤M, where T and M are the approximations given by the trapezoid and the midpoint rules respectively.

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Chapter 9.4, Problem 32E

Chapter 9.4, Problem 34E

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To prove: The area of shaded trapezoid in figure (a) is the same as the area of the shaded rectangle in figure (c), that is f ( x 1 ) Δ x

Solution Summary: The author proves that the area of the shaded trapezoid in figure (a) is the same as that of a right angled triangle. The tangent line at (x_

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To prove: The area of shaded trapezoid in figure (a) is the same as the area of the shaded rectangle in figure (c), that is f(x1)Δx

Why T≤∫abf(x)dx≤M, where T and M are the approximations given by the trapezoid and the midpoint rules respectively.

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