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To find the Integral of (sinxcos^2x) dx, we must first use our knowledge of integration and differentiation of simple trigonometric functions. Such as Sinx and Cosx. Combined with our knowledge of integra...

The first thing required is to find out what dy/dx is in terms of x. For this, we need to differentiate y with respect to x which be can so to each term of the polynomial. All you need to do is mutiply th...

∫(x³+x²+6)dx= ∫x³dx +∫x²dx+∫6dx =x⁴/4 +x³/3 +6x+C.

We can divide by (cosx)^2 across the identity (sinx)^2 + (cosx)^2 = 1 (which can be derived from properties of the unit circle and a bit of Pythagoras’ theorem) to achieve

[(sinx)^2 / (cosx)^2] + [...

P(J>2) = P(J=0)+P(J=1)     [split it up]

P(X=t)= (V^t)/t!*e^V       where V=4 in this case  [use the formula]

P(J>2) = 4^0/0!*e^4 + 4^1/1!*e^4

          =1/e^4 + 4/e^4  =  5e^-4...