Barrons AP Calculus

(b)

(c)

(d)

(e)

(f)

(g)

Since   .

.

For F(x)    to  equal   1,  we  need    a   region  under   f   whose   left    endpoint    is  4   with

area    equal   to  1.  The region  from    4   to  5   works   nicely; so      and x   =   10.

.

.

In  Figure  N6–22b  we  evaluate    the areas   in  the original    graph.

Figure  N6–22b

Measured from the lower limit of integration, 4, we have (with “f ” as an
abbreviation for “f(t) dt”):

We note that, since F ′(= f) is linear on (2,4), F is quadratic on (4,8); also, since f
′ is positive and increasing on (2,3), the graph of F is increasing and concave up
on (4,6), while since f ′ is positive and decreasing on (3,4), the graph of F is
increasing but concave down on (6,8). Finally, since F ′ is constant on (4,6), F is
linear on (8,12). (See Figure N6–22c.)