Mr. V Wiki Math - User contributions [en]

7.1 Integration By Parts/45

2022-11-30T00:30:09Z

Johnny C.:

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex]

&= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]

\end{align}

</math>

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]

&= \frac{3-1}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} [-cos(x) \bigg|_{0}^{\frac{\pi}{2}} \\[2ex]
&= \frac{2}{3} [-cos(\frac{\pi}{2}) -(-cos(0))] \\[2ex]
&= \frac{2}{3} [-1(0) -(-1)] \\[2ex]
&= \frac{2}{3}[1] \\[2ex]
&= \frac{2}{3} \\[2ex]

\end{align}

</math>

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^{2n+1} (x) dx \\[2ex]
&= \frac{2n+1-1}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n+1-2} (x) dx \\[2ex]
&= \frac{2n}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n-1} (x) dx \\[2ex]

\end{align}

</math>

<math>

\begin{align}

\int_{0}^{\frac{\pi}{2}} sin^5(x) dx &= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]
&= \frac{4}{5}\int_{0}^{\frac{\pi}{2}} (1-cos^2(x)) sin(x) dx \\[2ex]
&= -\int_{1}^{0} (1-u^2)\cdot du \\[2ex]
&= \frac{4}{5} \left[u - \frac{u^3}{3}\right]\bigg|_{0}^{1} \\[2ex]

& u= cos(x) \\[2ex]
& du= -sin(x) \\[2ex]
& dx= \frac{du}{-sin(x)} \\[2ex]
& \int sin^3(x) dx \\[2ex]
&= \frac{4}{5} [1-\frac{1}{3}] \\[2ex]
&= \frac{8}{15}

\end{align}

</math>

7.1 Integration By Parts/45

2022-11-29T05:52:02Z

Johnny C.:

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex]

&= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]

\end{align}

</math>

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]

&= \frac{3-1}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} [-cos(x) \bigg|_{0}^{\frac{\pi}{2}} \\[2ex]
&= \frac{2}{3} [-cos(\frac{\pi}{2}) -(-cos(0))] \\[2ex]
&= \frac{2}{3} [-1(0) -(-1)] \\[2ex]
&= \frac{2}{3}[1] \\[2ex]
&= \frac{2}{3} \\[2ex]

\end{align}

</math>

<math>

\begin{align}

&\int_{0}^{\frac{\pi}{2}} sin^5(x) dx \\[2ex]

&= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]
& u= cos(x) \\[2ex]
& du= -sin(x) \\[2ex]
& dx= \frac{du}{-sin(x)} \\[2ex]
& \int sin^3(x) dx \\[2ex]
&= \int (1-cos^2(x)) sin(x) dx \\[2ex]
&= \int (1-u^2)sin(x) \cdot \frac{du}{-sin(x)} \\[2ex]
&= -\int 1-u^2 du \\[2ex]
&= \frac{4}{5} [-u + \frac{1}{3}^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(x) + \frac{1}{3} cos^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(\frac{\pi}{2}) + \frac{1}{3} cos^3 (\frac{\pi}{2}) - (-cos(0) + \frac{1}{3} cos^3 (0)) \\[2ex]
&= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]

&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)] \\[2ex]
&= \frac{4}{5}[\frac{2}{3}] \\[2ex]
&= \frac{8}{15} \\[2ex]

\end{align}

</math>

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^{2n+1} (x) dx \\[2ex]
&= \frac{2n+1-1}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n+1-2} (x) dx \\[2ex]
&= \frac{2n}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n-1} (x) dx \\[2ex]

\end{align}

</math>

7.1 Integration By Parts/45

2022-11-29T05:48:48Z

Johnny C.: Created page with "<math> \begin{align} & \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex] &= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &..."

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex]

&= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]
&= \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex]

\end{align}

</math>

<math>

\begin{align}

& \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]

&= \frac{3-1}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex]
&= \frac{2}{3} [-cos(x) \bigg|_{0}^{\frac{\pi}{2}} \\[2ex]
&= \frac{2}{3} [-cos(\frac{\pi}{2}) -(-cos(0))] \\[2ex]
&= \frac{2}{3} [-1(0) -(-1)] \\[2ex]
&= \frac{2}{3}[1] \\[2ex]
&= \frac{2}{3} \\[2ex]

\end{align}

</math>

<math>

\begin{align}

&\int_{0}^{\frac{\pi}{2}} sin^5(x) dx \\[2ex]

&= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]
& u= cos(x) \\[2ex]
& du= -sin(x) \\[2ex]
& dx= \frac{du}{-sin(x)} \\[2ex]
& \int sin^3(x) dx \\[2ex]
&= \int (1-cos^2(x)) sin(x) dx \\[2ex]
&= \int (1-u^2)sin(x) \cdot \frac{du}{-sin(x)} \\[2ex]
&= -\int 1-u^2 du \\[2ex]
&= \frac{4}{5} [-u + \frac{1}{3}^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(x) + \frac{1}{3} cos^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(\frac{\pi}{2}) + \frac{1}{3} cos^3 (\frac{\pi}{2}) - (-cos(0) + \frac{1}{3} cos^3 (0)) \\[2ex]
&= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]

&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)]
&= \frac{4}{5}[\frac{2}{3}]
&= \frac{8}{15}

\end{align}

</math>

7.1 Integration By Parts/52

2022-11-28T05:26:36Z

Johnny C.: Created page with "<math> \begin{align} & \int{}^{} x^4 e^x &\int{}^{} x^n e^x dx = x^n e^x - n \int{}^{} x^{n-1} e^x dx & x^4 e^x - 4 \int{}^{} x^3 e^x dx \end{align} </math>"

<math>

\begin{align}

& \int{}^{} x^4 e^x
&\int{}^{} x^n e^x dx = x^n e^x - n \int{}^{} x^{n-1} e^x dx

& x^4 e^x - 4 \int{}^{} x^3 e^x dx

\end{align}
</math>

7.1 Integration By Parts/35

2022-11-28T05:22:23Z

Johnny C.: Created page with "<math> \begin{align} & \int_\sqrt{\frac{{\pi}}{2}}^\sqrt{\pi}\ \theta^3 cos(\theta^2) d\theta & u=\theta^2 \\[2ex] & du= 2\theta d\theta \\[2ex] & \frac{1}{2}du=\theta d\theta \end{align} </math> <math> \begin{align} & \int{}^{} u\cdot cos(u) du & w=u \\[2ex] & wu=du & dv= cos(u) dx & v= sin(u) \end{align} </math> <math> \begin{align} & u\cdot sin(u) - \int{}^{} sin(u) du &= \frac{1}{2}u \cdot sin(u) + \frac{1}{2}cos(u)\bigg|_{\frac{\pi}{2}}^{\pi} &= \fr..."

<math>

\begin{align}

& \int_\sqrt{\frac{{\pi}}{2}}^\sqrt{\pi}\ \theta^3 cos(\theta^2) d\theta

& u=\theta^2 \\[2ex]
& du= 2\theta d\theta \\[2ex]
& \frac{1}{2}du=\theta d\theta

\end{align}

</math>

<math>

\begin{align}

& \int{}^{} u\cdot cos(u) du
& w=u \\[2ex]
& wu=du
& dv= cos(u) dx
& v= sin(u)

\end{align}
</math>

<math>

\begin{align}

& u\cdot sin(u) - \int{}^{} sin(u) du
&= \frac{1}{2}u \cdot sin(u) + \frac{1}{2}cos(u)\bigg|_{\frac{\pi}{2}}^{\pi}
&= \frac{1}{2}(\pi) \cdot sin(\pi) + \frac{1}{2} cos(\pi) - (\frac{1}{2}(\frac{\pi}{2}) \cdot sin(\frac{\pi}{2}) + \frac{1}{2} cos(\frac{\pi}{2})) \\[2ex]
&= \frac{1}{2} (\pi)(0) + \frac{1}{2} (-1) - ( \frac{\pi}{4} + \frac{1}{2}(0) \\[2ex]
&= -\frac{1}{2} - \frac{\pi}{4}
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T19:21:20Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{green}\mathbf{y=x} \\
& \color{blue}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}-\frac{1}{4}x\right)dx = \int_{0}^{1} \left(\frac{3}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}\right)dx - \int_{1}^{2} \left(\frac{1}{4}x\right)dx

</math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln|x|\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2} \\[2ex]
&= \left[\frac{3(1)^2}{8}\right] + \left[ln|2|-ln|1|\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \frac{3}{8} + ln|2| - \frac{3}{8}
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T19:16:35Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{green}\mathbf{y=x} \\
& \color{blue}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}-\frac{1}{4}x\right)dx = \int_{0}^{1} \left(\frac{3}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}\right)dx - \int_{1}^{2} \left(\frac{1}{4}x\right)dx =
</math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln|x|\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln|2|-ln|1|\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \frac{3}{8} + ln|2| - \frac{3}{8}
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T19:08:11Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{green}\mathbf{y=x} \\
& \color{blue}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)\,dx + \int_{1}^{2} \left|(\frac{1}{x}-\frac{1}{4}x)\right|dx = \int_{0}^{1} \left|(\frac{3}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x})\right|dx - \int_{1}^{2} \left|(\frac{1}{4}x)\right|dx =
</math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln(x)\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= frac{3}{8} + ln(2)] - \frac{3}{8}
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T19:05:54Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{green}\mathbf{y=x} \\
& \color{blue}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)\,dx + \int_{1}^{2} \left|(\frac{1}{x}-\frac{1}{4}x)\right|dx = \int_{0}^{1} \left|(\frac{3}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x})\right|dx - \int_{1}^{2} \left|(\frac{1}{4}x)\right|dx =
</math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln(x)\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \left[\frac{3}{8}\right] + \left[ln(2)\right] - \left[\frac{3}{8}\right]
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T19:04:38Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{green}\mathbf{y=x} \\
& \color{blue}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left(x-\frac{1}{4}x\]right)\,dx + \int_{1}^{2} \left|(\frac{1}{x}-\frac{1}{4}x)\right|dx = \int_{0}^{1} \left|(\frac{3}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x})\right|dx - \int_{1}^{2} \left|(\frac{1}{4}x)\right|dx = </math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln(x)\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \left[\frac{3}{8}\right] + \left[ln(2)\right] - \left[\frac{3}{8}\right]
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T19:01:31Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{green}\mathbf{y=x} \\
& \color{blue}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left(x-\frac{1}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x}-\frac{1}{4}x)\right|dx = \int_{0}^{1} \left|(\frac{3}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x})\right|dx - \int_{1}^{2} \left|(\frac{1}{4}x)\right|dx = </math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln(x)\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \left[\frac{3}{8}\right] + \left[ln(2)\right] - \left[\frac{3}{8}\right]
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T18:59:28Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{green}\mathbf{y=x} \\
& \color{blue}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left|(x-\frac{1}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x}-\frac{1}{4}x)\right|d = \int_{0}^{1} \left|(\frac{3}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x})\right|dx - \int_{1}^{2} \left|(\frac{1}{4}x)\right|dx = </math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln(x)\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \left[\frac{3}{8}\right] + \left[ln(2)\right] - \left[\frac{3}{8}\right]
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-20T18:58:28Z

Johnny C.:

[[File:Desmos-graph100.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{royalblue}\mathbf{y=x} \\
& \color{green}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left|(x-\frac{1}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x}-\frac{1}{4}x)\right|d = \int_{0}^{1} \left|(\frac{3}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x})\right|dx - \int_{1}^{2} \left|(\frac{1}{4}x)\right|dx = </math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln(x)\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \left[\frac{3}{8}\right] + \left[ln(2)\right] - \left[\frac{3}{8}\right]
&= ln(2)
\end{align}
</math>

File:Desmos-graph100.png

2022-09-20T18:57:39Z

Johnny C.:

6.1 Areas Between Curves/27

2022-09-19T06:30:40Z

Johnny C.:

[[File:Desmos-graph.png|right|500px|]]

<math>
\begin{align}

& \color{red}\mathbf{y= \frac{1}{x} }
& \color{royalblue}\mathbf{y=x} \\
& \color{green}\mathbf{y= \frac{1}{4} x } \\
& x>0 \\

\end{align}

</math>

<math>\int_{0}^{1} \left|(x-\frac{1}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x}-\frac{1}{4}x)\right|d = \int_{0}^{1} \left|(\frac{3}{4}x)\right|dx + \int_{1}^{2} \left|(\frac{1}{x})\right|dx - \int_{1}^{2} \left|(\frac{1}{4}x)\right|dx = </math>

<math>
\begin{align}

&= \left[\frac{3x^2}{8}\right]\Bigg|_{0}^{1} + \left[\ln(x)\right]\Bigg|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg|_{1}^{2}=
&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
&= \left[\frac{3}{8}\right] + \left[ln(2)\right] - \left[\frac{3}{8}\right]
&= ln(2)
\end{align}
</math>

6.1 Areas Between Curves/27

2022-09-19T05:33:04Z

Johnny C.: Created page with "500px|"

[[File:Desmos-graph.png|right|500px|]]

5.5 The Substitution Rule/29

2022-09-04T04:19:02Z

Johnny C.: Created page with "<math> \int e^{tan(x)}sec^2(x)dx </math> <math> \begin{align} u &=tan(x) \\[2ex] du &= sec^2(x)dx \\[2ex] \end{align} </math> <math> \begin{align} \int e^udu \\[2ex] &= e^u+c \\[2ex] &= e^{tan(x)}+ c \end{align} </math>"

<math>
\int e^{tan(x)}sec^2(x)dx
</math>

<math>
\begin{align}

u &=tan(x) \\[2ex]
du &= sec^2(x)dx \\[2ex]

\end{align}
</math>

<math>
\begin{align}

\int e^udu \\[2ex]

&= e^u+c \\[2ex]
&= e^{tan(x)}+ c

\end{align}
</math>

5.5 The Substitution Rule/9

2022-09-04T04:06:08Z

Johnny C.:

<math>
\int (3x-2)^{20} dx
</math>

<math>
\begin{align}

u &=3x-2 \\[2ex]
du &= 3dx \\[2ex]
\frac{1}{3}du &= dx \\[2ex]

\end{align}
</math>

<math>
\begin{align}

\int u^{20}\cdot\frac{1}{3}du \\[2ex] &= \frac{1}{3}\int u^{20}du

&= \frac{1}{3}\cdot\frac{1}{20+1} u^{20+1}
&= \frac{1}{3}\cdot\frac{1}{21} u^{21}
&= \frac{1}{63} u^{21} + c
&= \frac{1}{63} (3x-2)^{21} + c

\end{align}
</math>

5.5 The Substitution Rule/9

2022-09-04T04:05:26Z

Johnny C.:

5.5 The Substitution Rule/9

2022-09-04T03:53:22Z

Johnny C.:

<math>
\int (3x-2)^{20} dx
</math>

<math>
\begin{align}

u &=3x-2 \\[2ex]
du &= 3dx \\[2ex]
\frac{1}{3}du &= dx \\[2ex]

\end{align}
</math>

<math>
\begin{align}

\int(u)^{20}/frac{1}{3}du \\[2ex]

&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex]
&= -\cos{(u)} + C \\[2ex]
&= -\cos{(\ln{(x)})} + C

\end{align}
</math>

5.5 The Substitution Rule/9

2022-09-04T03:42:34Z

Johnny C.:

<math>
\int (3x-2)^{20} dx
</math>

<math>
\begin{align}

u &=\ln(x) \\[2ex]
du &= \frac{1}{x}dx \\[2ex]

\end{align}
</math>

5.5 The Substitution Rule/9

2022-09-04T03:42:13Z

Johnny C.:

<math>
\int (3x-2)^(20) dx
</math>

<math>
\begin{align}

u &=\ln(x) \\[2ex]
du &= \frac{1}{x}dx \\[2ex]

\end{align}
</math>

5.5 The Substitution Rule/9

2022-09-04T03:42:00Z

Johnny C.:

<math>
\int (3x-2)^20 dx
</math>

<math>
\begin{align}

u &=\ln(x) \\[2ex]
du &= \frac{1}{x}dx \\[2ex]

\end{align}
</math>

5.5 The Substitution Rule/9

2022-09-04T03:41:45Z

Johnny C.: Created page with "<math> \int (3x-2)^20dx </math> <math> \begin{align} u &=\ln(x) \\[2ex] du &= \frac{1}{x}dx \\[2ex] \end{align} </math>"

<math>
\int (3x-2)^20dx
</math>

<math>
\begin{align}

u &=\ln(x) \\[2ex]
du &= \frac{1}{x}dx \\[2ex]

\end{align}
</math>

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:09:44Z

Johnny C.:

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:09:35Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\[2ex]
&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1-\frac{1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1-\frac{1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:09:13Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1-\frac{1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1-\frac{1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:09:03Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1-\frac{1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:08:14Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:07:51Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:07:30Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:07:06Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:06:47Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:06:24Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:06:10Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]

\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:05:52Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^-4\right)\,dt \\[2ex]

\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:05:39Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^-4\right)\,dt \\[2ex]
&= \frac{3t^{-4}{-3\bigg|_{1}^2 = -t^-3 \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:05:25Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}{-3\bigg|_{1}^2 = -t^-3 \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:05:01Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}{-3\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:04:33Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}{-3\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:03:32Z

Johnny C.:

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:03:21Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^{-4}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:02:59Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:02:35Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^{-3}\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:02:06Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\
&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]
&= 1+\frac{-1}{8} = \frac{7}{8}
\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:01:34Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:01:15Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\
&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:00:51Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(3t^{-4}\right)\,dt \\

\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:00:42Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(3t^{-4}\right)/,dt \\

\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}

5.3 The Fundamental Theorem of Calculus/25

2022-08-25T20:00:21Z

Johnny C.:

<math>
\begin{align}
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(3t^{-4}\,dt\right) \\

\end{align}
</math>

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1+\frac{-1}{8} = \frac{7}{8}