The Fundamental Theorem of Calculus

Evaluating an integral exactly as F(b) - F(a) through any antiderivative, and reading the integral of a rate as the total change it accumulates.

Active Calculus, The Fundamental Theorem of Calculus

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5 exercises

Activity 4.4.2

Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly. For each, sketch a graph of the integrand on the relevant interval and write one sentence that explains the meaning of the value of the integral in terms of the (net signed) area bounded by the curve.

$\int_{-1}^4 (2-2x) \, dx$
$\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx$
$\int_0^1 e^x \, dx$
$\int_{-1}^{1} x^5 \, dx$
$\int_0^2 (3x^3 - 2x^2 - e^x) \, dx$

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Activity 4.4.3

Use your knowledge of derivatives of basic functions to complete Table of antiderivatives. For each entry, your task is to find a function $F$ whose derivative is the given function $f$ . When finished, use the FTC and the results in the table to evaluate the three given definite integrals.

$\displaystyle \int_0^1 \left(x^3 - x - e^x + 2\right) \,dx$
$\displaystyle \int_0^{\pi/3} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt$
$\displaystyle \int_0^1 (\sqrt{x}-x^2) \, dx$

2.5

Exercise 13

When an aircraft attempts to climb as rapidly as possible, its climb rate (in feet per minute) decreases as altitude increases, because the air is less dense at higher altitudes. Given below is a table showing performance data for a certain single engine aircraft, giving its climb rate at various altitudes, where $c(h)$ denotes the climb rate of the airplane at an altitude $h$ .

Let a new function called $m(h)$ …

2.5

Exercise 14

In Chapter, we showed that for an object moving along a straight line with position function $s(t)$ , the object's “average velocity on the interval $[a,b]$ ” is given by .

More recently in Chapter, we found that for an object moving along a straight line with velocity function $v(t)$ , the object's “average value of its velocity function on $[a,b]$ ” is .

Are the “average velocity on the interval $[a,b]$ ” and the “average valu…

2.5

Exercise 15

In Table in Activity, we noted that for $x \gt 0$ , an antiderivative of $f(x) = \frac{1}{x}$ is $F(x) = \ln(x)$ . Here we observe that a key difference between $f(x)$ and $F(x)$ is that $f$ is defined for all $x \ne 0$ , while $F$ is only defined for $x \gt 0$ , and see how we can actually define an antiderivative of $f$ for all values of $x$ .

Suppose that $x \lt 0$ , and let $G(x) = \ln(-x)$ . Compute $G'(x)$ .
Explain why $G$ is an antiderivative of $f$ for $x \lt 0$ .
Let $H(x) = \ln(|x|)$ , and recall that . Explain why $H(x) = G(x)$ for $x \lt 0$ and $H(x) = F(x)$ …

2.5