integrate p(x-1/6L)-1/3*P*x dx from x=0 to x=L
Why isn't this question working on WolframAlpha for the definite integral with respect to x on the interval [0,L] of function (p(x-1/6L)-1/3Px)?
Hi, Try this one integrate p (x-1/6L)-1/3*P x dx from x=0 to x=L
Sometimes a little Mathematica notation can coax WolframAlpha into providing an answer
Integrate[p (x-1/6 L)-1/3*P*x, {x,0,L}]
or here
