§5.1 Constructing accurate graphs of antiderivatives

Introduction

Constructing the graph of an antiderivative

Multiple antiderivatives of a single function

Functions defined by integrals

A function $F$ has derivative $f$ whose graph shows $f$ positive on $(0,2)$ and $(5,7)$, and negative on $(2,5)$. Then $F$ is

For the same $F$ with $F' = f$, the concavity of $F$ follows the monotonicity of $f$, so $F$ is concave up where $f$ is increasing. If $f$ increases on $(0,1)$ and $(4,6)$ and decreases on $(1,3)$ and $(6,7)$, then $F$ is

With $F' = f$ positive on $(0,2)$, negative on $(2,5)$, and positive again on $(5,7)$, the function $F$ has a relative maximum at

A function $f$ has piecewise linear derivative $f'$ that is positive on $\left(0,\frac{3}{2}\right)$, negative on $\left(\frac{3}{2},4\right)$, positive on $(4,6)$, and zero outside $[0,6]$. The function $f$

For $g(x) = |x| - 1$ with the condition $G(-1) = 0$, an antiderivative $G$ is built from quadratic pieces. On $x \lt 0$ the correct piece is

For $h(x) = \sin(x)$ with the condition $H(0) = 1$, the antiderivative is

A function $p$ equals $x^2$ on $(0,1)$, $-(x-2)^2$ on $(1,2)$, and $0$ elsewhere, with $P(0) = 1$. Since the area under $x^2$ on $(0,1)$ is $\frac{1}{3}$, the value $P(1)$ equals

For $A(x) = \int_1^x g(t)\,dt$ where $g$ is positive on $(0,1.5)$ and $(4,6)$ and negative on $(1.5,4)$, the function $A$

For the same $A$, the integrand $g$ increases on $(0,1)$ and $(3,5)$ and decreases on $(1,3)$ and $(5,6)$. The concavity of $A$ is

With $A$ increasing on $(0,1.5)$ and $(4,6)$ and decreasing on $(1.5,4)$, the function $A$ has

To sketch the complete graph of $y = F(x)$ from the earlier work in this lesson, the correct final assembly is

For an integrand $f$ that vanishes for all $t \lt 0$ and $t \gt 7$, with $F(0) = -1$ and accumulated signed areas giving $F(7) = \frac{3\pi}{4} - \frac{5}{2}$, the values $F(-1)$ and $F(8)$ are

To sketch the complete graph of $y = A(x)$, which accumulates $g$ with $A(0) = -\frac{1}{2}$, the correct approach is

For $A(x) = \int_1^x g(t)\,dt$ and $B(x) = \int_0^x g(t)\,dt$ with $\int_0^1 g(t)\,dt = \frac{1}{2}$, we can write $B(x) = A(x) + c$ for a constant $c$. Find $c$ as a decimal.

Using the table values $A(0) = -\frac{1}{2}$, $A(1) = 0$, $A(2) = 0$, $A(3) = -2$, $A(4) = -3.5$, $A(5) = -2$, compute the signed area of $g$ on $[4,5]$.

With $F(0) = -1$ and signed area $\int_0^1 f(t)\,dt = \frac{1}{2}$, compute $F(1)$ as a decimal.

If $F$ and $G$ are both antiderivatives of the same continuous function $f$ on an interval, then $F(x) - G(x)$ is constant, so changing the initial condition only shifts the graph vertically.