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 2 years ago
\[\frac{1}{2}\int_{b1/2}^{b+1/2}\frac{1}{\sqrt{x}}dx=\frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{x}}dx=\sqrt{1/2b}+\sqrt{1/2+b}\]
 2 years ago
\[\frac{1}{2}\int_{b1/2}^{b+1/2}\frac{1}{\sqrt{x}}dx=\frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{x}}dx=\sqrt{1/2b}+\sqrt{1/2+b}\]

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Accipiter46
 2 years ago
Best ResponseYou've already chosen the best response.0How can I format this to fit the window?

Accipiter46
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\int_{b1/2}^{b+1/2}\frac{1}{\sqrt{x}}dx=\] \[\frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{x}}dx=\] \[\sqrt{1/2b}+\sqrt{1/2+b}\]

TheViper
 2 years ago
Best ResponseYou've already chosen the best response.0Yes I too have the same question :/ @satellite73 @sasogeek @ParthKohli Plz help :)

Accipiter46
 2 years ago
Best ResponseYou've already chosen the best response.0I simply broke the equation up in three parts manually :)

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.5\[\begin{aligned} \frac{1}{2}\int_{b1/2}^{b+1/2}\frac{1}{\sqrt{x}}dx &= \frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{x}}dx \\ &=\sqrt{1/2b}+\sqrt{1/2+b} \end{aligned}\]There's no way to get it to format automatically. However, you can use the environment `\begin{aligned}` and `\end{aligned}` to format lines in regards to one another and make it look pretty nice. The code I used to make the above math was `\begin{aligned}` `\frac{1}{2}\int_{b1/2}^{b+1/2}\frac{1}{\sqrt{x}}dx` `&= \frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx` `+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{x}}dx \\` `&=\sqrt{1/2b}+\sqrt{1/2+b}` `\end{aligned}` "\\" marks a new line, and the "&" signs are automatically lined up.

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1\[\begin{aligned} \frac 12\int\limits_{b1/2}^{b+1/2}\frac{dx}{\sqrt{x}} &= \frac12\int\limits_{b1/2}^{0}\frac{dx}{\sqrt{x}} +\frac12\int\limits_{0}^{b+1/2}\frac{dx}{\sqrt{x}}\\ &=\sqrt{1/2b}+\sqrt{1/2+b} \end{aligned}\] \[\begin{aligned} \frac 12\int\limits_{b1/2}^{b+1/2}\frac{dx}{\sqrt{x}} &= \frac12\int\limits_{b1/2}^{0}\frac{dx}{\sqrt{x}} +\frac12\int\limits_{0}^{b+1/2}\frac{dx}{\sqrt{x}}\\ &=\sqrt{1/2b}+\sqrt{1/2+b} \end{aligned }\]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1@KingGeorge ``` ``` create multiline here ``` ```

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1``` \begin{aligned} \frac 12\int\limits_{b1/2}^{b+1/2}\frac{dx}{\sqrt{x}} &=\frac 12\int\limits_{b1/2}^{0}\frac{dx{\sqrt{x}} +\frac12\int\limits_{0}^{b+1/2}\frac{dx}{\sqrt{x}}\\ &=\sqrt{1/2b}+\sqrt{1/2+b}\end{aligned} ``` ah, pretty colours
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