∫4 1 𝑔(𝑥)𝑑𝑥 = −8, g(1) = −3, and 𝑔(4) = 6 Solve: ∫4 1 𝑥∗𝑔′(𝑥)dx

Answer 1

IBP

U = x dV = g'(x)

dU = dx V = g(x)

the integral is

x g(x) - integral [ g(x) dx]

limit of x g(x) as x->4 is 4 * 6 = 24

limit of x g(x) as x->1 is 1 * -3 = -3

definite integral is given to be -8

therefore, [24 - -3] - -8

=27 - -8

= 35