Question

y = (1/4)x^2 - (1/2)lnx..over the interval (1, 7e) ...what is the arc length ? need help pleease

Answer 1

\[\int \text{Sqrt}\left[1+F'[x]^2\right]\text{dx}\]

F(x)=\[\frac{1}{4}x^2-\frac{1}{2}\text{Log}[x]\]

F'(x)=\[-\frac{1}{2 x}+\frac{x}{2}\]

F'(x)^2=\[\frac{1-2 x^2+x^4}{4 x^2}\]

Answer 2

91.7389 usin s= int( sqrt(1+(y prime)^2) 1,7e)

Answer 3

\[\int_1^{7 e} \sqrt{\frac{x^4+4 x^2-2 x^2+1}{4 x^2}} \, dx\]

Answer 4

ya i got that far imranmeah

Answer 5

\[\int\limits_1^{7 e} \sqrt{\frac{x^4+2 x^2+1}{4 x^2}} \, dx\]
\[(x^2 + 1)^2/4 x^2
\]

\[\left(\frac{\left(x^2+1\right)}{2x}\right)^2\]

Answer 6

\[\int_1^{7 e} \sqrt{\left(\frac{\left(x^2+1\right)}{2x}\right)^2} \, dx\]

Answer 7

\[\int_1^{7 e} \left(\frac{\left(x^2+1\right)}{2x}\right) \, dx\]

easy U-Sub from here

Answer 8

Sorry U-Sub won't work
\[\int\limits_1^{7 e} \left(\frac{\left(x^2\right)}{2x}\right) \, dx\text{ }+\int\limits_1^{7 e} \left(\frac{(1)}{2x}\right) \, dx\]

\[\int_1^{7 e} \frac{x}{2} \, dx\text{ }+\int_1^{7 e} \frac{1}{2x} \, dx\]

Answer 9

\[\frac{1}{2}\int _1^{7 e}x dx\text{ }+\frac{1}{2}\int_1^{7 e} \frac{1}{x} \, dx\]

Answer 10

It is basic integral from here

Answer 11

(1/2)+(1/2)(((x^2)/2)+lnx))?

Answer 12

\[\frac{1}{2}\int\limits _1^{7 e}x dx\text{ }+\frac{1}{2}\int\limits_1^{7 e} \frac{1}{x} \, dx\]

\[\frac{1}{4}x^2\text{from} \{1,7E\}\text{ }+\frac{1}{2}\text{ }\ln [x]\{1,7e\}\]
\[\left(\frac{1}{4}49E^2-\frac{1}{4}\right)+\frac{1}{2}\text{Log}[7E]\]

Answer 13

thanks yo u

Answer 14

thanks yo u