Let

Find each value and represent each value using a graph of the function (1 – *t*).

(a) *F*(3)

(b) *F*(1)

(c) *F*(0)

(d) *F*(-1)

(e) *F*(-2)

Answer

(a)

On the graph of (1 – *t*), *F*(3) is the weighted area between (1 – *t*) and the *t*-axis on the interval [-1,3]. Since this weighted area consists of two triangles of equal sizes, one above the axis and one below, it makes sense that *F*(3) should be 0.

(b)

On the graph of (1 – *t*), *F*(1) is the weighted area between (1 – *t*) and the *t*-axis on the interval [-1,1]. This area is a triangle with base 2 and height 2, thus area 2.

(c)

On the graph of (1 – *t*), *F*(0) is the weighted area between (1 – *t*) and the *t*-axis on the interval [-1,0].

(d)

Since the upper and lower limits of integration are the same,

When we try to represent *F*(-1) on the graph of (1 – *t*) we get a line, which has no area.

(e)

The area between the graph of (1 – *t*) and the *t*-axis on the interval [-2,-1] lies above the *x*-axis. Since we're integrating from right to left, we count this area negatively: