Mathematics

Problem

Define a piecewise function

h(x)=\begin{cases}\sqrt{x}, & 0\le x\le4,\\4, & 4< x\le6,\end{cases}

and let

k(x)=x-2

for

0\le x\le6

. Region

T

consists of all points in the plane that lie between the graphs of

h

and

k

on the closed interval

[0,6]

. (a) Determine all

x

–values in

[0,6]

for which

h(x)=k(x)

. (b) For each subinterval determined by the

x

–values found in part (a), indicate which function lies above the other. Justify your answer with appropriate reasoning. (c) Write an expression, involving one or more definite integrals, whose value equals the area of region

T

. Do not evaluate yet. (d) Evaluate the integral expression from part (c) exactly, and then give a decimal approximation correct to the nearest tenth.

Intuitive explanation

The key is to recognize that region $T$ is bounded by two different curves, and you need to figure out where each one forms the upper and lower boundary. Start by sketching both functions on $[0,6]$ : $h(x)$ is a square root curve that flattens to a horizontal line, while $k(x)$ is a simple linear function. Find their intersection points to determine which function is on top in each interval, then set up integrals to sum the vertical "slices" of area between them.

Intuitive explanation

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