Pattab

2022-07-01

Determine antiderivative of f.

So I have a function:

$f(t)=\{\begin{array}{ll}-1,& \text{if}-2\le t\le 0\\ 2-\sqrt{{t}^{2}-6t+9},& \text{if}0t6\\ -1,& \text{if}6\le t\le 8\end{array}$

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: $2-|t-3|$.

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.

So I have a function:

$f(t)=\{\begin{array}{ll}-1,& \text{if}-2\le t\le 0\\ 2-\sqrt{{t}^{2}-6t+9},& \text{if}0t6\\ -1,& \text{if}6\le t\le 8\end{array}$

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: $2-|t-3|$.

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.

Leslie Rollins

Beginner2022-07-02Added 25 answers

Step 1

First let's be explicit about what you say you already figured out:

$f(t)=\{\begin{array}{ll}-1,& \text{if}-2\le t\le 0\\ t-1,& \text{if}0t\le 3\\ 5-t,& \text{if}3t6\\ -1,& \text{if}6\le t\le 8\end{array}$

Step 2

$F(x)=\{\begin{array}{rlrl}& {\int}_{-2}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt& & \text{if}-2\le x\le 0,\\ & {\int}_{-2}^{0}f(t)\phantom{\rule{thinmathspace}{0ex}}dt+{\int}_{0}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt& & \text{if}0\le x\le 3,\\ & {\int}_{-2}^{0}f(t)\phantom{\rule{thinmathspace}{0ex}}dt+{\int}_{0}^{3}f(t)\phantom{\rule{thinmathspace}{0ex}}dt+{\int}_{3}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt& & \text{if}3\le x\le 6,\\ & \text{and so on.}\end{array}$

Note that here I've written $\text{\u201c}\le \text{''}$ rather than $\text{\u201c}<\text{''}$ since altering the value of f at isolated points does not affect the integral.

First let's be explicit about what you say you already figured out:

$f(t)=\{\begin{array}{ll}-1,& \text{if}-2\le t\le 0\\ t-1,& \text{if}0t\le 3\\ 5-t,& \text{if}3t6\\ -1,& \text{if}6\le t\le 8\end{array}$

Step 2

$F(x)=\{\begin{array}{rlrl}& {\int}_{-2}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt& & \text{if}-2\le x\le 0,\\ & {\int}_{-2}^{0}f(t)\phantom{\rule{thinmathspace}{0ex}}dt+{\int}_{0}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt& & \text{if}0\le x\le 3,\\ & {\int}_{-2}^{0}f(t)\phantom{\rule{thinmathspace}{0ex}}dt+{\int}_{0}^{3}f(t)\phantom{\rule{thinmathspace}{0ex}}dt+{\int}_{3}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt& & \text{if}3\le x\le 6,\\ & \text{and so on.}\end{array}$

Note that here I've written $\text{\u201c}\le \text{''}$ rather than $\text{\u201c}<\text{''}$ since altering the value of f at isolated points does not affect the integral.

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