Question

I ask myself "How do I make a stretchy (evaluation) bar?"

Answer 1

?

Answer 2

A stretchy evaluation bar is useful in evaluating a definite integral using the fundamental theorem of calculus. Never mind that. Here is how to make it:

\( \huge \left. {1 \over y} \right|_{x=3}^4 \)


\ ( \huge \left. {1 \over y} \right|_{x=3}^4 \ )

\left.
use \left period for the left delimiter. Period makes it invisible. Use \right with keyboard | for the right delimiter. This makes the bar stretchy. \left and \right make stretchy delimiters, but they have to be paired.

Not stretchy:

\( \huge {1 \over y} |_{x=3}^4 \)

Answer 3

Stretchy but not huge
\( \left. {1 \over y} \right|_{x=3}^4 \)

Not stretch, not huge
\( {1 \over y} |_{x=3}^4 \)

Answer 4

The/A Fundamental Theorem of Calculus

\(\Large \displaystyle{ F^\prime(x) = f(x) \iff \int_a^b f(x)dx = F(b) - F(a) = \left. F(x) \right|_a^b } \)

Answer 5

wow

Answer 6

\( \left. \dfrac{1} {y} \right|_{x=3}^4\)

Answer 7

Another interesting one to use is limits.

\(\sum_{i=1}^5i^2 \) vs \(\sum\limits_{i=1}^5i^2 \)

Answer 8

Howeeverl, limits does not work with all things. For example, your evaluationbar example:

\(\left.\dfrac{1}{y} \right|\limits_{x=3}^4\)

But if I make it an integral:

\(\int_{x=3}^4\dfrac{1}{y} \) vs \(\int \limits_{x=3}^4\dfrac{1}{y} \)

Oh, and the rules also change for block vs. inline mode. `\( bla \)` does inline. `\[ bla \]` does block.

\[\int_{x=3}^4\dfrac{1}{y} \] vs \[\int \limits_{x=3}^4\dfrac{1}{y} \]

Now, because I have been using dfrac, not just frac, the fractions are larger.

Answer 9

Hmm. I tend to use \over for fractions instead of frac and have not noticed a difference. I've been using \displaystyle because I only yesterday learned about square brackets.

Answer 10

Example of longhand multiplication. This may be better spacing or your mileage may vary. use \; and \: to make minor adjustments.

\[\Large \displaystyle \begin{align} x &+ h \cr x &+ h \cr \hline hx &+ h^2 \cr x^2 +\;\;\; hx& \cr\hline x^2 + 2hx &+ h^2 \cr x& + h \cr \hline hx^2 +2h^2x& + h^3 \cr x^3 +2hx^2 +\;\;\;h^2x& \cr \hline x^3 + 3hx^2 +3h^2x &+ h^3 \end{align}\]

Answer 11

Definite integral

\[\large\int_{a}^{b} f(x) dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\Delta x\]

Answer 12

You can also use \!, the negative space, to adjust things, or \hspace and a very precise value.
\(a+b\)
\(a\!+\!b\)

Answer 13

Linear (classical) motion cheat sheet:

\[ \large \int adt = at + C \]

The antiderivative of acceleration is velocity with respect to time. The constant \(C = v_0\), the initial velocity.

The dimension of acceleration is length / (time x time). The integral of acceleration with respect to time (i.e velocity) therefore has dimensions of length/ time.

\[ \large \int (at + v_0)dt = {1 \over 2}at^2 + v_0t + C \]

The antiderivative of velocity with respect to time is displacement. s is often used for displacement. Do not mistake this s for "speed." Use d in your own work to avoid confusion. In this case \(C = s_0 \) the original displacement

The dimension of displacement is length.

Going the other way:

\[ \large {d \over {dt}} s(t) = v(t) \]
\[ \large {d \over {dt}} v(t) = a \]

Answer 14

\(\cfrac[r]{right}{justified}\) in a fraction. \cfrac[r]