The Fundamental Theorem of Calculus Exercises

Answer

(a)

Because we're integrating to the left, this value is the negative of the weighted area between sin t and the t-axis on [-π, 0]. So it's the positive value of this area:

(b)

This makes sense, because the weighted areas between sin t and the t-axis on [0, 2π] cancel out:

(c)

Since we're integrating to the left, this is the negative of the weighted area between sin t and the t-axis on :

The weighted area on  is zero, because the areas above and below the axis on this interval are equal:

We're left with the negative of the negatively weighted area between sin t and the axis on , which is this area:

(d) We're integrating starting at 0. When we integrate sin t on the interval [0, 0] we get 0, so

When x is any integer multiple of 2π (positive or negative),

because the weighted areas the integral accumulates cancel each other out:

For any value of x in (0,π], the function

accumulates a positively weighted area, so the value of the function will be positive:

For any value of x in (π, 2π), the area below the t-axis is smaller than the area above the t-axis, so the function

is still positive:

Since , when we move x into the interval (2π, 3π] the values of F(x) are still positive:

And similarly for x in the interval (3π, 4π):

We can generalize to say that F(x) is zero when x is an integer multiple of 2π and positive for all positive values of x that aren't multiples of 2π.

When x is a negative multiple of 2π, we know that F(x) = 0. What about when x < 0 is not a multiple of 2π?

When x < 0 the function F(x) integrates to the left, so we count areas below the axis positively and areas above the axis negatively.

For x in [-π, 0) the function F(x) counts this area positively. This means F(x) > 0 for x in [-π, 0).

For x in (-2π, -π) the function F(x) considers the area below the axis positively, and the area above the axis negatively. Since there's more area below the axis than above the axis, F(x) is positive.

Generalizing, F(x) will be positive for all x < 0 that aren't multiples of 2π.

Putting all our observations together, we can finally answer the question.

(i) F(x) is positive for all x that aren't multiples of 2π.

(ii) F(x) is never negative.

(iii) F(x) is zero for all x that are multiples of 2π.

In the previous integral we used 0 as the lower limit of integration, but we can build a similar function using any lower limit of integration we want.

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:

Answer

The most obvious one is

We could also make

or

and so on and so forth. Since e^ex is defined everywhere, we can use any constant for the lower limit of integration when building the antiderivative.