The Second Fundamental Theorem of Calculus Examples

Let f(x) = sin x and a = 0. Define

Find

(a) F(π)

(b) 

(c) 

To find the value F(x), we integrate the sine function from 0 to x.

(a) To find F(π), we integrate sine from 0 to π:

This means we're accumulating the weighted area between sin t and the t-axis from 0 to π:

Evaluating the integral, we get

The value of F(π) is the weighted area between sin t and the horizontal axis from 0 to π, which is 2.

(b) To find  we substitute  in for x:

On the graph, we're accumulating the weighted area between sin t and the t-axis from 0 to .

The value 1 makes sense as an answer, because the weighted areas

and

cancel each other out:

We're left with one-half the area

Since we found that

,

it makes sense for

to be 1.

(c) To find  we put in  for x. The upper limit of integration  is less than the lower limit of integration 0, but that's okay.

You can write the answer as

if you want.

On the graph, we're accumulating the weighted area between sin t and the t-axis from 0 to . This means we're integrating going left:

Since we're accumulating area below the axis, but going left instead of right, it makes sense to get a positive number for an answer.

Let . Find each value and represent each value using a graph of the function 2t.

(a) F(2)

(b) F(1)

(c) F(0)

(d) F(-1)

(e) F(-2)

(a)

The value F(2) is this area:

(b) Since we're integrating over an interval of length 0,

When we try to represent this on a graph, we get a line, which has no area:

(c)

Since we're integrating to the left, F(0) is the negative of this area:

(d) Since 2t is an odd function,

The areas above and below the t-axis on [-1,1] are the same:

(e)

The weighted area between 2t and the t-axis on [-1,1] is 0, so we're left with the area on [-2,1]. Since we're integrating to the left and this area is below the t-axis, we count this area positively: