Question

Bounds on an integral Suppose ƒ is continuous on [a, b] with ƒ''(𝓍) > 0 on the interval. It can be shown that (b―a) ƒ [(a + b) /2] ≤ ∫ₐᵇ ƒ(𝓍) d𝓍 ≤ (b―a) [ (ƒ(a) + ƒ(b)) /2]

(a) Assuming ƒ is nonnegative on [a, b], draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions. b.

Accepted Answer

Step 1: Understand the given inequality. Since $$ f''(x) \u003e 0  on$$ $$[a, b]$$, the function $$ f  is $$convex on this interval. The inequality states that:

$$
(b - a) f\left(\frac{a + b}{2}\right) \leq \int_a^b f(x) \, dx \leq (b - a) \frac{f(a) + f(b)}{2}
$$

This means the integral of $$ f $$ over $$[a, b] is $$bounded below by the area of a rectangle with height $$ f  at $$the midpoint, and above by the area of a trapezoid formed by the values at the endpoints. Step 2: Sketch the graph of $$ f  on $$the interval $$[a, b]. $$Since $$ f  is $$convex, the curve lies below the line segment connecting $$ (a, f(a)) $$ and $$ (b, f(b)) . $$Mark the points $$ (a, f(a)) $$, $$ (b, f(b)) $$, and the midpoint $$ \left(\frac{a+b}{2}, f\left(\frac{a+b}{2}\right)\right) . $$Step 3: Illustrate the lower bound: draw a rectangle with base $$ (b - a) $$ and height $$ f\left(\frac{a+b}{2}\right) . $$This rectangle represents the quantity $$ (b - a) f\left(\frac{a+b}{2}\right) $$, which is less than or equal to the integral. Step 4: Illustrate the upper bound: draw the trapezoid formed by the points $$ (a, 0) $$, $$ (a, f(a)) $$, $$ (b, f(b)) $$, and $$ (b, 0) . $$The area of this trapezoid is $$ (b - a) \frac{f(a) + f(b)}{2} $$, which is greater than or equal to the integral. Step 5: Discuss conclusions: Since $$ f  is $$convex and nonnegative, the integral (area under the curve) lies between the area of the midpoint rectangle and the trapezoid formed by the endpoints. This reflects how convexity controls the shape of $$ f $$ and provides useful bounds for the integral.

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Geometric Interpretation of Integral Inequalities