Question

How to evaluate the integral of ∫[x- (1/2x)]^2 dx

Answer 1

you can expand

Answer 2

then +/- of integral

Answer 3

is your question written like this one. \[\huge ∫[x- \frac{1}{2x}]^2\]
@justine97

Answer 4

most likely

Answer 5

Why did you put evaluate if you have no bounds? o.O
that's just an indefinite integral, if you're asking for that just put what is the antiderivative of..
xD

Answer 6

you can evaluate without bounds you just put + C

Answer 7

he just forgot to put dx. honest mistake

Answer 8

\[\int\limits(x^2-1+\frac{ 1 }{ 4x^2 })dx\]

Answer 9

\[\huge ∫[x- \frac{1}{2x}]^2 dx =\huge ∫[x^2+ (\frac{1}{2x})^2 -2 \times x \times \frac{1}{2x}]\] dx
\[ =∫[x^2+ \frac{1}{4x^2} -1] dx = ∫x^2 dx + \int\limits \frac{1}{4x^2} dx -\int\limits 1dx\]
\[= ∫x^2 dx + \frac{1}{4} \int\limits x^{-2} dx -\int\limits 1dx\]
\[= \frac{x^3}{3} + \frac{1}{4} (\frac{x^{-1}}{-1}) -x +c\]
\[= \frac{x^3}{3} - (\frac{x^{-1}}{4}) -x +c\]
\[= \frac{x^3}{3} - \frac{1}{4x} -x +c\]
@justine97

Answer 10

\[\frac{ x^3 }{ 3 }-x+\frac{ 1 }{ 4 }\frac{ x^{-1} }{ -1 } + C\]
just simplify

Answer 11

That guy actually took the time to write all of it in LaTeX xD

Answer 12

@AshleyAlley<3 thank you

Answer 13

ya I wouldn't do that unless it's a complicated demonstration

Answer 14

but our student left us without a word or hint of life