Question

Integral of .75x(1-x^2)dx from -1 14 years ago

Answer 1

Wolfram keeps saying 1, my integrating keeps saying 0.

Answer 2

So the integral is \[\int\limits_{-1}^{1}x[.75(1-x ^{2})]dx\]

Answer 3

Which becomes... \[.75((x^{2}/2)-(x^{4}/4)) from -1 \to 1\]

Answer 4

\[.75([((1)^{2}/2)-((1)^{4}/4)]-[((-1)^{2}/2)-((-1)^{4}/4)])\]

Answer 5

\[.75[(1/4)-(1/4)]\]

Answer 6

.75[0] = 0 What's wrong with my integral logic?

Answer 7

you just use u sub

u = 1-x^2

du= -2x dx

Answer 8

\[.75x(1-x^2) = .75x-.75 x^3\]

Answer 9

\[\frac{.75}{2} \int (u) \, du\]

Answer 10

adjust the limit of integration accordingly

Answer 11

\[\int .75x -.75x^3\ dx=\frac{.75}{2}x^2-\frac{.75}{4}x^4\]

Answer 12

\[\frac{.75}{2}-\frac{.75}{4}-\frac{.75}{2}+\frac{.75}{4}=0\]

Answer 13

unless stated otherwise, just let the integration take care of itself and the signed areas will cancel if they cancel

Answer 14

\[\frac{1}{2} (-0.75) \int_0^0 u \, du\]

Answer 15

\[\int\limits_{-1}^{1}\frac{3}{4}x(1-x^2) dx=\frac{3}{4} \int\limits_{-1}^{1}(x-x^3) dx\]
\[\frac{3}{4}(\frac{x^2}{2}-\frac{x^4}{4})|_{-1}^{1}=\frac{3}{4}[(\frac{1}{2}-\frac{1}{4})-(\frac{1}{2}-\frac{1}{4})]\]
\[\frac{3}{4}[0]=0\]
and i like imran's way better less work

Answer 16

http://www.wolframalpha.com/input/?i=integral+of+.75x%281-x^2%29+from+-1+to+1

Answer 17

wolfram say 0 too

Answer 18

imram way is snappy. adjust the limits, forget the anti-derivative