Mr. V Wiki Math - User contributions [en]

7.1 Integration By Parts/51

2022-11-29T03:55:56Z

Elizabethh58413: Created page with "<math> \text{use Exercise 47 to evaluate} \int(\ln{x})^3dx </math> <math> \text{Exercise 47:}\qquad x(\ln{x})^n-n\int(\ln{x})^{n-1}dx </math> <math> \int(\ln{x})^3dx= x(\ln{x})^3-3\int(\ln{x})^2dx </math> <math> u = (\ln{x})^2 \qquad dv = dx </math> <math> du = 2\frac{1}{x}\ln{x}dx \qquad v = x </math> <math> \begin{align} \int(\ln{x})^3dx &= x(\ln{x})^3-3\int(\ln{x})^2dx = x(\ln{x})^3-3[x(\ln{x})^2 - 2\int(\ln{x})dx] \\[2e..."

<math> \text{use Exercise 47 to evaluate} \int(\ln{x})^3dx </math> 
<math> \text{Exercise 47:}\qquad x(\ln{x})^n-n\int(\ln{x})^{n-1}dx </math> 
<math>
\int(\ln{x})^3dx= x(\ln{x})^3-3\int(\ln{x})^2dx </math> 
<math> u = (\ln{x})^2 \qquad dv = dx </math> 
<math> du = 2\frac{1}{x}\ln{x}dx \qquad v = x </math> 

<math>
\begin{align}
\int(\ln{x})^3dx &= x(\ln{x})^3-3\int(\ln{x})^2dx = x(\ln{x})^3-3[x(\ln{x})^2 - 2\int(\ln{x})dx] \\[2ex]
&= x(\ln{x})^3-3[x(\ln{x})^2 -2(x\ln{x} - \int1dx) = x(\ln{x})^3-3[x(\ln{x})^2 -2x\ln{x} + 2x] \\[2ex]
& =x(\ln{x})^3 - 3x(\ln{x})^2+ 6x\ln{x} - 6x +c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/37

2022-11-29T03:09:35Z

Elizabethh58413:

<math> \int x\ln({x})dx </math> 
<math>
\begin{align}
u &= x + 1 \\[2ex]
x &= u-1 \\[2ex]
du &= dx \\[2ex]
\end{align}
</math> 
<math>\int(u-1)\ln({u}) du </math> 
<math> w = \ln{u} \qquad dv = (u-1)dv </math> 
<math> du = \frac{1}{u}du \qquad v = \frac{1}{2}u^2-u </math> 
<math>
\begin{align}
\int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u - \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]
&= \ln{u} \cdot \frac{1}{2}u^2 - u - \int(\frac{1}{2}u - 1)du \\[2ex]
&= \ln{u} \cdot \frac{1}{2}u^2 - u - [\frac{1}{4} u^2 - u] + c \\[2ex]
&=\ln({1+x})(\frac{1}{2}(x^2+2x+1)-(1+x)) - (\frac{1}{4}(1+x)^2-(1+x)) + c \\[2ex]
&= \ln({1+x})(\frac{1}{2}x^2 + x + \frac{1}{2} - x- 1) - (\frac{1}{4}(1+x)^2-(1+x)) + c \\[2ex]
&= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 - \frac{1}{2}x-\frac{1}{4}+1+x+c \\[2ex]
&= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 + \frac{1}{2}x + \frac{3}{4} + c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/37

2022-11-29T02:58:28Z

Elizabethh58413:

<math> \int x\ln({x})dx </math> 
<math>
\begin{align}
u &= x + 1 \\[2ex]
x &= u-1 \\[2ex]
du &= dx \\[2ex]
\end{align}
</math> 
<math>
\begin{align}
\int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u - \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]
&= \ln{u} \cdot \frac{1}{2}u^2 - u - \int(\frac{1}{2}u - 1)du \\[2ex]
&= \ln{u} \cdot \frac{1}{2}u^2 - u - [\frac{1}{4} u^2 - u] + c \\[2ex]
&=\ln({1+x})(\frac{1}{2}(x^2+2x+1)-(1+x)) - (\frac{1}{4}(1+x)^2-(1+x)) + c \\[2ex]
&= \ln({1+x})(\frac{1}{2}x^2 + x + \frac{1}{2} - x- 1) - (\frac{1}{4}(1+x)^2-(1+x)) + c \\[2ex]
&= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 - \frac{1}{2}x-\frac{1}{4}+1+x+c \\[2ex]
&= \ln{(1+x})(\frac{1}{2}x^2-\frac{1}{2}) - \frac{1}{4}x^2 + \frac{1}{2}x + \frac{3}{4} + c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/37

2022-11-29T02:57:44Z

Elizabethh58413:

7.1 Integration By Parts/37

2022-11-29T02:37:22Z

Elizabethh58413: Created page with "<math> \int x\ln({x})dx </math> <math> \begin{align} u &= x + 1 \\[2ex] x &= u-1 \\[2ex] du &= dx \\[2ex] \end{align} </math> <math> \begin{align} \int(u-1)\ln({u}) du &= \ln{u} \cdot \frac{1}{2}u^2 - u \int(\frac{1}{2}u^2-u) \cdot \frac{1}{u} du \\[2ex]"

7.1 Integration By Parts/43

2022-11-29T02:16:00Z

Elizabethh58413:

<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> 
<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math> 
<math>
\begin{align}

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
\end{align}
</math>

<math> \text{Now evaluate}\int\sin^4(x)dx </math>

<math>
\begin{align}
\int\sin^4(x)dx &= - \frac{1}{4}\cos(x)\sin^3(x) + \frac{3}{4} \int\sin^2(x)dx \\[2ex]
&= - \frac{1}{4}\cos(x)\sin^3(x) + \frac{3}{4}(-\frac{1}{4}\sin(2x) + \frac{x}{2}) \\[2ex]
&= - \frac{1}{4}\cos(x)\sin^3(x) - \frac{3}{16}\sin(2x) + \frac{3}{8}x + c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/43

2022-11-29T02:13:21Z

Elizabethh58413:

7.1 Integration By Parts/43

2022-11-29T00:26:18Z

Elizabethh58413:

7.1 Integration By Parts/43

2022-11-29T00:25:05Z

Elizabethh58413:

<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> 
<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>
<math>
\begin{align}

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/43

2022-11-29T00:24:46Z

Elizabethh58413:

<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math>
<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>
<math>
\begin{align}

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/43

2022-11-29T00:23:44Z

Elizabethh58413:

<math>
\begin{align}

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/43

2022-11-29T00:23:05Z

Elizabethh58413:

<math>
\begin{align}

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]

\end{align}
</math>

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]

&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

7.1 Integration By Parts/43

2022-11-29T00:22:28Z

Elizabethh58413:

<math>
\begin{align}
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x +

\end{align}
</math>

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]

&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

7.1 Integration By Parts/43

2022-11-29T00:22:15Z

Elizabethh58413:

<math>
\begin{align}
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot

\end{align}
</math>

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]

&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

7.1 Integration By Parts/43

2022-11-29T00:21:55Z

Elizabethh58413:

<math>
\begin{align}
\int\sin^2(x)dx

\end{align}
</math>

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]

&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

7.1 Integration By Parts/43

2022-11-29T00:21:36Z

Elizabethh58413:

<math>
\begin{align}

\end{align}
</math>

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]

&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

7.1 Integration By Parts/43

2022-11-29T00:21:22Z

Elizabethh58413:

<math>
\begin{align}
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]

\end{align}
</math>

&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

7.1 Integration By Parts/43

2022-11-29T00:20:27Z

Elizabethh58413:

<math>
\begin{align}
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
\end{align}
</math>

7.1 Integration By Parts/43

2022-11-29T00:17:31Z

Elizabethh58413:

\begin{align}
<math>

\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]

</math>
\end{align}

7.1 Integration By Parts/43

2022-11-29T00:16:46Z

Elizabethh58413:

<math>
/begin{align}
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]
/end{align}
</math>

7.1 Integration By Parts/43

2022-11-28T23:58:42Z

Elizabethh58413:

7.1 Integration By Parts/43

2022-11-28T23:57:25Z

Elizabethh58413:

7.1 Integration By Parts/43

2022-11-28T23:55:37Z

Elizabethh58413:

7.1 Integration By Parts/43

2022-11-28T23:54:57Z

Elizabethh58413:

7.1 Integration By Parts/43

2022-11-28T21:21:35Z

Elizabethh58413:

7.1 Integration By Parts/43

2022-11-28T21:18:40Z

Elizabethh58413: Created page with " <math> \int{sin^2x}dx = - \frac{1}{2}cos(x) \cdot sin^{2-1}x + \frac{2-1}{2} \int sin^0xdx </math> "

6.1 Areas Between Curves/13

2022-09-20T20:12:40Z

Elizabethh58413:

<math>\color{red}{y=12-x^2}</math>

<math>\color{blue}{y=x^2-6}</math>

[[File:Screen Shot 2022-09-20 at 1.07.09 PM.png|right|500px]]

<math>\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx </math>

<math>
\begin{align}
12-x^2 &= x^2-6 \\
18 &= 2x^2 \\
9 &=x^2 \\
\pm3 &= x
\end{align}
</math>

<math>
\begin{align}
\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx &= [(12x-\frac{1}{3}x^3) - (\frac{1}{3}x^3 - 6x)] \Bigg|_{-3}^{3} \\
& = [(12(3) - \frac{1}{3}(3)^3) - (\frac{1}{3}(3)^3-6(3) )] - [(12(-3) - \frac{1}{3}(-3)^3) - (\frac{1}{3}(-3)^3-6(-3) )] \\
& = 36-(-36) \\
& = 72
\end{align}
</math>

File:Screen Shot 2022-09-20 at 1.07.09 PM.png

2022-09-20T20:09:23Z

Elizabethh58413:

6.1 Areas Between Curves/13

2022-09-20T20:05:45Z

Elizabethh58413:

<math>\color{red}{y=12-x^2}</math>

<math>\color{blue}{y=x^2-6}</math>

<math>\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx </math>

<math>
\begin{align}
12-x^2 &= x^2-6 \\
18 &= 2x^2 \\
9 &=x^2 \\
\pm3 &= x
\end{align}
</math>

<math>
\begin{align}
\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx &= [(12x-\frac{1}{3}x^3) - (\frac{1}{3}x^3 - 6x)] \Bigg|_{-3}^{3} \\
& = [(12(3) - \frac{1}{3}(3)^3) - (\frac{1}{3}(3)^3-6(3) )] - [(12(-3) - \frac{1}{3}(-3)^3) - (\frac{1}{3}(-3)^3-6(-3) )] \\
& = 36-(-36) \\
& = 72
\end{align}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:58:51Z

Elizabethh58413:

<math>\color{red}{y=12-x^2}</math>

<math>\color{blue}{y=x^2-6}</math>

<math>\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx </math>

<math>
\begin{align}
12-x^2 &= x^2-6 \\
18 &= 2x^2 \\
9 &=x^2 \\
\pm3 &= x
\end{align}
</math>

<math>
\begin{align}
\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx &= (12x-\frac{1}{3}x^3) - (\frac{1}{3}x^3 - 6x) \Bigg|_{-3}^{3} \\
& = 36-(-36) \\
& = 72
\end{align}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:58:20Z

Elizabethh58413:

<math>\color{red}{y=12-x^2}</math>

<math>\color{blue}{y=x^2-6}</math>

<math>\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx </math>
<math>
\begin{align}
12-x^2 &= x^2-6 \\
18 &= 2x^2 \\
9 &=x^2 \\
\pm3 &= x
\end{align}
</math>

<math>
\begin{align}
\int_{-3}^{3} \left|(12-x^2) - (x^2-6) \right| dx &= (12x-\frac{1}{3}x^3) - (\frac{1}{3}x^3 - 6x) \Bigg|_{-3}^{3} \\
& = 36-(-36) \\
& = 72
\end{align}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:37:12Z

Elizabethh58413:

<math>\color{red}{y=12-x^2}</math>

<math>\color{blue}{y=x^2-6}</math>

<math>\int_{3}^{-3} \left|(12-x^2) - (x^2-6) \right| dx </math>

6.1 Areas Between Curves/13

2022-09-20T19:34:25Z

Elizabethh58413:

<math>\color{red}{y=12-x^2}</math>

<math>\color{red}{y=x^2-6}</math>

6.1 Areas Between Curves/13

2022-09-20T19:32:56Z

Elizabethh58413:

<math>
\begin{align}
&\color{red}{y=12-x^2}
&\color{blue}{y=x^2-6}\\
&x=-3
&x=3 \\
\end{align}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:32:38Z

Elizabethh58413:

<math>
\begin{align}
&\color{red}{y=12-x^2}
&\color{blue}{y=x^2-6}\\
&x=-3
&x=3 \\
\end{align}

6.1 Areas Between Curves/13

2022-09-20T19:25:57Z

Elizabethh58413:

<math>\color{red}{y=12-x^2}</math>

<math>\color{blue}{y=x^2-6}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:24:56Z

Elizabethh58413:

<math>

\color{red}{y=12-x^2}
\color{blue}{y=x^2-6}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:23:04Z

Elizabethh58413:

<math>

\color{red}{y=12-x^2}\hspace{1cm}
\color{blue}{y=x^2-6}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:22:50Z

Elizabethh58413:

<math>

\color{red}{y=12-x^2}\hspace{1cm}\color{blue}{y=x^2-6}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:18:56Z

Elizabethh58413:

<math>

\color{red}{y=12-x^2}
\color{blue}{y=x^2-6}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:18:01Z

Elizabethh58413:

<math>

\color{red}{y=12-x^2}
\color{blue}{x^2-6}
</math>

6.1 Areas Between Curves/13

2022-09-20T19:13:05Z

Elizabethh58413: Created page with "<math> \color{red}\mathbf{y=12-x^2} \color{blue}\mathbf{x^2-6}"

<math>

\color{red}\mathbf{y=12-x^2}
\color{blue}\mathbf{x^2-6}

5.5 The Substitution Rule/59

2022-09-06T02:04:54Z

Elizabethh58413:

<math> \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx </math> 

<math>
\begin{align}
u &=\frac{1}{x} \\[2ex]
du &=-\frac{1}{x^2}dx \\[2ex]
-du &=\frac{1}{x^2}dx \\[2ex]
\end{align}
</math>

New upper limit:<math>\frac{1}{2} = \frac{1}{(2)} </math> 
New lower limit: <math> 1 = \frac{1}{(1)} </math>

<math>
\begin{align}
\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx)
&=\int_{1}^{\frac{1}{2}}e^u\,(-du) \\[2ex]
&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]
&=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex]
&=-\sqrt{e} - (-e^1) \\[2ex]
&=e-\sqrt{e}
\end {align}
</math>

5.5 The Substitution Rule/59

2022-09-06T02:02:20Z

Elizabethh58413:

5.5 The Substitution Rule/59

2022-09-06T02:01:32Z

Elizabethh58413:

<math> \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx </math> 

<math>
\begin{align}
u &=\frac{1}{x} \\[2ex]
du &=-\frac{1}{x^2}dx \\[2ex]
-du &=\frac{1}{x^2}dx \\[2ex]
\end{align}
</math>

New upper limit:<math>\frac{1}{2} = \frac{1}{(2)} </math> 
New lower limit: <math> 1 = \frac{1}{(1)} </math>

<math>
\begin{align}
\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx)
&=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex]
&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]
&=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex]
&=e-\sqrt{e}
\end {align}
</math>

5.5 The Substitution Rule/59

2022-09-06T01:58:01Z

Elizabethh58413: Created page with "<math> \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx </math> <math> \begin{align} u &=\frac{1}{x} \\[2ex] du &=-\frac{1}{x^2}dx \\[2ex] -du &=\frac{1}{x^2}dx \\[2ex] \end{align} </math> New upper limit:<math> \frac{1}{2} = \frac{1}{(2)} <\math> New lower limit: <math> 1 = \frac{1}{(1)} <\math> <math> \begin{align} \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx) &=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex] &=-\int_{1}..."

<math> \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx </math> 

<math>
\begin{align}
u &=\frac{1}{x} \\[2ex]
du &=-\frac{1}{x^2}dx \\[2ex]
-du &=\frac{1}{x^2}dx \\[2ex]
\end{align}
</math>

New upper limit:<math> \frac{1}{2} = \frac{1}{(2)} <\math> 
New lower limit: <math> 1 = \frac{1}{(1)} <\math>

<math>
\begin{align}
\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx)
&=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex]
&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]
&=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex]
&=e-\sqrt{e}
\end {align}
</math>

5.3 The Fundamental Theorem of Calculus/11

2022-08-25T19:24:02Z

Elizabethh58413: Created page with "<math> g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt </math> <math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>"

<math> g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt </math> 
<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> 
<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>

Elizabeth

2022-08-23T19:51:45Z

Elizabethh58413: Created page with "<math>\cos^2\theta + sin^2\theta = 1</math>"

<math>\cos^2\theta + sin^2\theta = 1</math>

Test Here

2022-08-23T15:57:19Z

Elizabethh58413:

Use this page to test commands. Do not delete this line.

<math>f(x)=\frac{1}{x}</math>

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