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So I have a function:

$$f (t) = \begin{cases} -1, & \text{if } {-2} \le t \le 0 \\ 2 - \sqrt{t^2-6t+9}, & \text{if } 0 < t < 6 \\ -1, & \text{if } 6 \le t \le 8 \\ \end{cases}$$

I need to find the antiderivative of $f(t)$ on the interval: $-2 < t < 8$

And I need to investigate whether the antiderivative has a maximum value on the interval.

So far I've figured out that the second function $2 - \sqrt{t^2-6t+9}$ can be described as an absolute value: $2 - \vert{t-3}\vert$.

That makes it 4 different functions in total that describes $f(t)$

However, I'm not sure exactly how to formulate a function for the antiderivative on the given interval.

Michael Hardy
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PiotreX
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2 Answers2

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Hint. Draw a picture of $f$. You should see a graph made up of four straight line segments. Look at the set of all antiderivative for each segment and adjust the constants to that the antiderivatives match where the segments meet.

Ethan Bolker
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First let's be explicit about what you say you already figured out: $$f (t) = \begin{cases} -1, & \text{if } {-2} \le t \le 0 \\ t-1, & \text{if } 0<t\le3 \\ 5-t, & \text{if } 3 < t < 6 \\ -1, & \text{if } 6 \le t \le 8 \\ \end{cases}$$

$$ F(x) = \left\{ \begin{align} & \int_{-2}^x f(t) \, dt & & \text{if } {-2}\le x \le 0, \\[4pt] & \int_{-2}^0 f(t)\,dt + \int_0^x f(t)\,dt & &\text{if } 0\le x \le 3, \\[4pt] & \int_{-2}^0 f(t)\,dt + \int_0^3 f(t)\,dt + \int_3^x f(t)\,dt & & \text{if } 3\le x \le 6, \\[4pt] & \text{and so on.} \end{align} \right. $$ Note that here I've written $\text{“}{\le}\text{''}$ rather than $\text{“}{<}\text{''}$ since altering the value of $f$ at isolated points does not affect the integral.

Michael Hardy
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