4.9: Antiderivatives

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

An antiderivative of a function $f(x)$ is a function $F(x)$ such that $F'(x) = f(x)$ .

If $F(x)$ is a function such that $F'(x) = f(x)$ , then a general antiderivative of $f(x)$ is $F(x)+C$ .

Let $F(x)$ denote a particular antiderivative of $f(x)$ .

(Power Rule) If $f(x) = x^n$ and $n \neq -1$ , then $F(x) = \frac {1}{n+1}x^{n+1}$ .
(Logs) If $f(x) = \frac {1}{x}$ , then $f'(x) = \ln |x|$ .
(Exponentials) If $f(x) = b^x$ with $b>0$ , then $F(x) = \frac {1}{\ln (b)}b^x$ .

Consider $f(x) = x^2+e^x$ .

Find a general antiderivative of $f(x)$ .
We can add the antiderivatives of $x^2$ and $e^x$ together to get the antiderivative of $f(x)$ . By the power rule above, the antiderivative of $x^2$ is $\answer {x^3/3}$ (without the $+C$ ) and the antiderivative of $e^x$ is $\answer {e^x}$ . When we add them, we get a particular antiderivative of $\answer {x^3/3+e^x}$ . Therefore, the general antiderivative is $F(x) = \answer {x^3/3+e^x+C}$ .
Find the antiderivative $F(x)$ of $f(x)$ satisfying $F(0) = 2$ .
By the previous part, a general antiderivative is $F(x) = \frac {x^3}{3}+e^x+C.$ We want to find $C$ satisfying $F(0) = 2$ . Plugging in $x=0$ gives $F(0) = \answer {1+C}.$ Since we want $F(0) = 2$ , we set this expression equal to $2$ to get $C = \answer {1}$ . Therefore, the specific antiderivative going through the point $(0,2)$ is $F(x) = \answer {\frac {x^3}{3}+e^x+1}.$

The general antiderivative of $f(x) = \cos (2x)$ is $F(x) = \answer {\frac {1}{2}\sin (2x)+C}$ .

Find the antiderivative $F(x)$ of each of the following functions going through the specified point.

$f(x) = 2e^x$ with $F(0) = 1$ : $F(x) = \answer {2e^x-1}$ .
$f(x) = \frac {1}{x} + \cos (x-1)$ with $F(1) = 1$ : $F(x) = \answer {\ln |x| +\sin (x-1) + 1}$ .
$f(x) = \frac {x^2+x}{x^4}$ with $F(1) = 0$ : $F(x) = \answer {-\frac {1}{x}-\frac {1}{2x^2}+\frac {3}{2}}$ .

A ball is thrown downward from a height of $100$ ft with an initial speed of $5$ ft/second. Find the height $h$ of the ball as a function of time. You may use without justification that the acceleration is $a(t) = -32$ ft/s $^2$ .

Note that $a(t) = h''(t)$ . To get to $h$ , we will need to take two antiderivatives, so the process is a bit longer. If we take one antiderivative of acceleration, we will get velocity, so let’s do that first. The general antiderivative of $-32$ is $\answer {-32t+C}$ . We know the initial speed is $5$ , but since the rock was thrown downward, the initial velocity is $v(0) = \answer {-5}$ . Plugging in $t=0$ into the expression for velocity and setting it equal to this initial velocity gives $C = \answer {-5}$ . Therefore, $v(t) = \answer {-32t-5}.$ We now take another antiderivative to get our height function. The general antiderivative of $v(t) = -32t-5$ is $\answer {-16t^2-5t+C}$ . We know the initial height is $h(0) = \answer {100}$ . Plugging in $t=0$ into the general antiderivative and setting it equal to $100$ gives $C= \answer {100}$ . Therefore, $h(t) = \answer {-16t^2-5t+100}.$

A rock is thrown upward from ground level and it hits the ground $2$ seconds later. How fast upward was the rock thrown? $\answer {32}$ feet/second.