5.5 The Substitution Rule/54: Difference between revisions
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(Created page with "<math> \int_{0}^{7} \sqrt{4+3x}dx </math> <math> \begin{align} u &=\ln(x) \\[2ex] du &= \frac{1}{x}dx \\[2ex] \end{align} </math> <math> \begin{align} \int \frac{\sin{(\ln{(x))}}}{x}dx &= \int\frac{1}{x}\sin(\ln{(x)})dx = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[2ex] &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} </math>") |
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<math> | <math> | ||
\int_{0}^{ | \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 &= \frac{1}{ | du &= 2xdx \\[2ex] | ||
\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}^{\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
New upper limit:
New lower limit: