5.5 The Substitution Rule/54: Difference between revisions
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<math> | <math> | ||
\int_{0}^{\sqrt{\pi}} x\cos{x^2}\,dx | \int_{0}^{\sqrt{\pi}} x\cos{(x^2)}\,dx | ||
</math> | </math> | ||
| Line 7: | Line 7: | ||
\begin{align} | \begin{align} | ||
u &= | u &=x^2 \\[2ex] | ||
du &= | du &= 2xdx \\[2ex] | ||
\frac{1}{ | \frac{1}{2}du &= xdx \\[2ex] | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
New upper limit: <math>\pi = (\sqrt{\pi})^2</math><br> | |||
New lower limit: <math>0 = (0)^2</math> | |||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
\int_{0}^{ | \int_{0}^{\sqrt{\pi}} x\cos{(x^2)}\,dx &= \int_{0}^{\sqrt{\pi}} (xdx)\cos{(x^2)} \\[2ex] | ||
&= \ | &= \int_{0}^{\pi} \left(\frac{1}{2}du\right)\cos{(u)} = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}du \\[2ex] | ||
&= | &= \frac{1}{2}\sin{(u)}\bigg|_{0}^{\pi} \\[2ex] | ||
&= | &= \frac{1}{2}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex] | ||
&= 0 | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 22:50, 28 August 2022
![{\displaystyle {\begin{aligned}u&=x^{2}\\[2ex]du&=2xdx\\[2ex]{\frac {1}{2}}du&=xdx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/18c922fd60a28dedb46a2819a6f7ab99b233dc17)
New upper limit:
New lower limit:
![{\displaystyle {\begin{aligned}\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx&=\int _{0}^{\sqrt {\pi }}(xdx)\cos {(x^{2})}\\[2ex]&=\int _{0}^{\pi }\left({\frac {1}{2}}du\right)\cos {(u)}={\frac {1}{2}}\int _{0}^{\pi }\cos {(u)}du\\[2ex]&={\frac {1}{2}}\sin {(u)}{\bigg |}_{0}^{\pi }\\[2ex]&={\frac {1}{2}}\sin {(\pi )}-{\frac {1}{2}}\sin {(0)}\\[2ex]&=0\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d30fc1f61658d5bb276f432ee3bb7718a814f5c9)