<img src="https://tex.z-dn.net/?f=%20%5Csf%20Evaluate%5Cint%20log%20%5C%3A%20x%20%5C%3A%20dx" id="TexFormula1" title=" \sf Evalu

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2 years ago

ate\int log \: x \: dx" alt=" \sf Evaluate\int log \: x \: dx" align="absmiddle" class="latex-formula">
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Thenkewww ☃️

Mathematics

2 answers:

Greeley [361]2 years ago

8 0

Answer:

x(log(x) - 1) + C

Step-by-step explanation:

The formula for the derivative of a logarithm function is:

$\frac{d}{dx} log_{a}(x) = \frac{1}{xlog(a)}$

For your case, the base is 10. Now that we have the derivative, lets use integration by parts. I will be differentiating log(x) and integrating 1:

u = log(x)

du = dx/xlog(10)

dv = 1

v = x

You get:

log(x)x - integral(dx/log(10))

Therefore the answer is:

log(x)x - x/log(10) + C

x(log(x) - 1) + C

Georgia [21]2 years ago

6 0

$\sf \huge \underline \pink{ \purple {Answer : }}$

$\: \:$

$\: \: \: \: \: \: \: \rm \large log \: xdx = \int \: log \: x *\: 1dx$

$\:$

$\: \: \: \: \: \: \: \rm \large = log \: x \int 1 \: dx - \int \: ( \int \: 1dx \: \frac{d}{dx} log \: x)dx$

$\: \:$

$\: \: \: \: \: \: \: \rm \large \: = x \: log \: x - x + c$

$\: \: \:$

$\: \: \: \: \: \: \: \rm \large \: = log \: x \: (x) - \int \: x \: \frac{1}{x} dx$

$\: \:$

$\: \: \: \: \: \: \: \rm \large = x \: log \: x - \int \: 1dx$

$\: \:$

$\: \: \: \: \: \: \ \rm \color{red} \bold{\large = x(log \: x - 1) + c}$

$\: \: \: \: \: \: \: \:$

$\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\$

Hope Helps!

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use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 ), (15 comma

lyudmila [28]

Answer:

volume V of the solid

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

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We then add up the volumes of all these slices

$\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}$

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$\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2$

But

$\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)$

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so we have, after replacing and simplifying, the sum of the slices equals

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Now we take the limit when n tends to infinite (the slices get thinner and thinner)

$\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}$

and the volume V of our solid is

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

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