Newton Leibnitz formula
If h(x) and g(x) are differentiable functions of x then,
ddx ∫h(x)g(x) f(t)dt = f[h(x)].h'(x) – f[g(x)].g'(x)
Example : Evaluate ddt ∫t3t2 1logx dx
Solution : We have,
ddt ∫t3t2 1logx dx = 1logt3 ddt (t3) – 1logt2 ddt (t2)
= 3t23logt – 2t2logt = t(t–1)logt