Question

Evaluate the definite integral.

function: (t+8)(t^2+3) with respect to variable t
lower limit: -sqrt(2)
upper limit: sqrt(2)

Answer 1

∫[-√2,√2] (t+8)(t^2+3) dt
= ∫[-√2,√2] t^3 + 8t^2 + 3t + 24 dt
= 1/4 t^4 + 8/3 t^3 + 3/2 t^2 + 24t [-√2,√2]
= 176/3 √2

Answer 2

Thank you!