Verification

[Erik Lehnsherr]

Are these correct? Did I miss anything?


$\displaystyle \text{1.\,Prove that}\,\,$ $\displaystyle \int_0^2 x^2 \sqrt{(x^2 + 4)}\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2})$

$\displaystyle u = x^2 + 4\,\,\,\,\,\,\,\,\,u - 4 = x^2$

\(\displaystyle = \int \frac{(u - 4)u^\frac{1}{2}}{2(u - 4)^\frac{1}{2}}\,\dot\,du\,\,\,\,\,\,\,\,\,du = 2x\,dx\)
$\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{du}{dx} = dx$

\(\displaystyle = \frac{1}{2}\int\frac{ (u - 4)^1}{u - 4^\frac{1}{2}}\,\dot\,\,u^\frac{1}{2}\)

$\displaystyle = \frac{1}{2} \int \frac{u^\frac{3}{2} - 4u^\frac{1}{2}}{(u - 4)}$




$\displaystyle \text{2.\,\,Show that} \int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \ln\bigg(\frac{32}{11}\bigg)$
$\displaystyle \int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \frac{A}{(x - 6)} + \frac{B}{(x + 1)}$

$\displaystyle Ax + A + Bx - 6B$
$\displaystyle A + B = 0, A - 6b = 7$
$\displaystyle A = -B\,\,\,\,\,(-B) - 6B = 7$
$\displaystyle \therefore A = 1\,\,\,\,\,-7B = 7\,\,\,B = -1$

$\displaystyle \frac{1}{(x - 6)} + \frac{1}{x + 1}$

$\displaystyle \bigg[\ln|x - 6| - \ln|x + 1|\bigg]_7^{10} + C$

$\displaystyle \ln |\frac{x - 6}{x + 1}| - \ln |\frac{x - 6}{x + 1}|$
$\displaystyle \ln |\frac{4}{11}| - \ln |\frac{1}{8}|$

$\displaystyle \ln \bigg(\frac{\frac{4}{11}}{\frac{1}{8}}\bigg)$

$\displaystyle \ln\bigg(\frac{4}{11} \times 8\bigg) = \ln \bigg(\frac{32}{11}\bigg)$

 

[tkhunny]

I have trouble believing that that much work doesn't result in a little confidence.

1) Does't quite appear to be done.

2a) Your second equation is entirely incorrect. Where did the integral go? Integral = A/stuff + B/stuff
2b) "+C" is not correct.
2c) Why did you write this wonderfully long name for zero (0)? ln|(x-6)/(x+1)| - ln|(x-6)/(x+1)|

Think about what you are doing. Remember basic rules.

1) NEVER write one of these "=" unless you mean it.
2) Don't write things that are meaningless.
3) If you are working with just part of a problem, please label it so.

Good work. Clean them up and finish.

 

[lookagain]

Are these correct? Did I miss anything?


$\displaystyle \text{1.\,Prove that}\,\,$ $\displaystyle \int_0^2 x^2 \sqrt{(x^2 + 4)}\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2})$

$\displaystyle u = x^2 + 4\,\,\,\,\,\,\,\,\,u - 4 = x^2$

\(\displaystyle = \int \frac{(u - 4)u^\frac{1}{2}}{2(u - 4)^\frac{1}{2}}\,\dot\,du\,\,\,\,\,\,\,\,\,du = 2x\,dx\)


\(\displaystyle = \frac{1}{2}\int\frac{ (u - 4)^1}{(u - 4)^\frac{1}{2}}\,\dot\,\,u^\frac{1}{2}du\)

$\displaystyle = \frac{1}{2} \int \frac{u^\frac{3}{2} - 4u^\frac{1}{2}}{(u - 4)}du$




$\displaystyle \text{2.\,\,Show that} \int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \ln\bigg(\frac{32}{11}\bigg)$

$\displaystyle \frac{7}{x^2 - 5x - 6}\,\, = \frac{A}{(x - 6)} + \frac{B}{(x + 1)}$

$\displaystyle Ax + A + Bx - 6B$
$\displaystyle A + B = 0, A - 6b = 7$
$\displaystyle A = -B\,\,\,\,\,(-B) - 6B = 7$
$\displaystyle \therefore A = 1\,\,\,\,\,-7B = 7\,\,\,B = -1$

$\displaystyle \frac{1}{(x - 6)} - \frac{1}{x + 1}$

$\displaystyle \int_7^{10}\bigg[ \frac{1}{x - 6} - \frac{1}{x + 1}\bigg]dx =$

$\displaystyle \bigg[\ln|x - 6| - \ln|x + 1|\bigg]_7^{10} =$

$\displaystyle \ln \bigg|\frac{10 - 6}{10 + 1}\bigg| - \ln \bigg|\frac{7 - 6}{7 + 1}\bigg|=$

$\displaystyle \ln \bigg|\frac{4}{11}\bigg| - \ln \bigg|\frac{1}{8}\bigg|=$

$\displaystyle \ln \bigg(\frac{\frac{4}{11}}{\frac{1}{8}}\bigg)=$

$\displaystyle \ln\bigg(\frac{4}{11} \cdot \frac{8}{1}\bigg) =$

$\displaystyle \ln \bigg(\frac{32}{11}\bigg)$

The above are more revisions, such as deletions, adding parentheses/"dx,"
adding in equal signs, and changing other symbols.

 

[Erik Lehnsherr]

Are these correct? Did I miss anything?


$\displaystyle \text{1.\,Prove that}\,\,$ $\displaystyle \int_0^2 x^2 \sqrt{(x^2 + 4)}\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2})$

$\displaystyle u = x^2 + 4\,\,\,\,\,\,\,\,\,u - 4 = x^2$

\(\displaystyle = \int \frac{(u - 4)u^\frac{1}{2}}{2(u - 4)^\frac{1}{2}}\,\dot\,du\,\,\,\,\,\,\,\,\,du = 2x\,dx\)


\(\displaystyle = \frac{1}{2}\int\frac{ (u - 4)^1}{(u - 4)^\frac{1}{2}}\,\dot\,\,u^\frac{1}{2}du\)

$\displaystyle = \frac{1}{2} \int \frac{u^\frac{3}{2} - 4u^\frac{1}{2}}{(u - 4)}du$



$\displaystyle \text{2.\,\,Show that} \int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \ln\bigg(\frac{32}{11}\bigg)$

$\displaystyle \frac{7}{x^2 - 5x - 6}\,\, = \frac{A}{(x - 6)} + \frac{B}{(x + 1)}$

$\displaystyle Ax + A + Bx - 6B$
$\displaystyle A + B = 0, A - 6b = 7$
$\displaystyle A = -B\,\,\,\,\,(-B) - 6B = 7$
$\displaystyle \therefore A = 1\,\,\,\,\,-7B = 7\,\,\,B = -1$

$\displaystyle \frac{1}{(x - 6)} - \frac{1}{x + 1}$

$\displaystyle \int_7^{10}\bigg[ \frac{1}{x - 6} - \frac{1}{x + 1}\bigg]dx =$

$\displaystyle \bigg[\ln|x - 6| - \ln|x + 1|\bigg]_7^{10} =$

$\displaystyle \ln \bigg|\frac{10 - 6}{10 + 1}\bigg| - \ln \bigg|\frac{7 - 6}{7 + 1}\bigg|=$

$\displaystyle \ln \bigg|\frac{4}{11}\bigg| - \ln \bigg|\frac{1}{8}\bigg|=$

$\displaystyle \ln \bigg(\frac{\frac{4}{11}}{\frac{1}{8}}\bigg)=$

$\displaystyle \ln\bigg(\frac{4}{11} \cdot \frac{8}{1}\bigg) =$

$\displaystyle \ln \bigg(\frac{32}{11}\bigg)$

The above are more revisions, such as deletions, adding parentheses/"dx,"
adding in equal signs, and changing other symbols.

Thank you.