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Explain why all solutions of the equation 3sinx = x - 1 must lie between x=-2 and x=4.

1. 2 ∫ x^3 - x dx (i expanded the orginal equation) -1 = [x^4/4 - x^2/2] = [4 - 2] - [1/4 - 1/2] = 4 1/4 sq. units hmm...my answer is diff from urs...maybe i got it wrong... 2. for this question...it wud rili help if u draw the graph first for the equation... when y= 0, x = 0, 1...

yep...i did use cos x= 1 - t^2 1 + t^2 using that i came up with this.... 2∫ 1 - t^2 dt (1 + 3t^2)(1+t^2) kinda stuck wid that at the moment....any hints to continue wid that?

using the substiution t= tan x/2 how do you integrate this: cos x 2 - cos x thnx!:)

whoops...pressed sumthing... newayz to continue... Testing n = 1 LHS = 1 RHS = 1/6 x 1 x (1+1)x(2+1) = 1 ...therefore true for n=1 let n=K 1^2+2^2+...k^2 = 1/6 k (k+1)(2k+1) show the result is true for n=k+1 1^2+2^2...k^2+(k+1)^2 = 1/6 (k+1)(k+2)(2k+3) LHS = 1/6 k...

Show that 1^2+2^2+....+n^2 = 1/6 n(n+1)(2n+1) for n=1, 2, 3, Testing n=1 LHS = 1 RHS = 1/6 x 1 x (1+1) x (2+1)