**Stewart Calculus Problems Plus Answers**

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*Suppose f(1)=f’(1)=0, f” is continuous on [0,1], abs(f”(x)) is less than or = to 3 for all x . Show that abs(integral of f(x) from 0 to 1) is less than or = to 1/2.*

(Taken from Stewart’s Calculus Early Transcendentals 6th Edition Chapter 7 Problems Plus)

This does not appear to be true:

Intuitively, you could define a piecewise function(with continuous derivative bounded by +/-3)

with four pieces:

the line y = 3x

a circular arc segment with small radius

a line segment with slope -3

another circular arc with slope 0, at x = 1

Judiciously constructed, the integral, or area under the curve would certainly be greater than 1/2

since the area of the left half is very nearly 1/2

do you actually have the solution manual for the problems plus? email me at [email protected] if you do. thanks!