Evaluate the following definite integrals

Question

a. ∫₀⁴ t³-3t dt

b. ∫_-pi/2^pi sin(x) dx

Answer 1

integral of t^3 -3t

=t^4/4 -3t^2/2

evaluated from 0 to 4

= 4^4/4 - 3(4^2)/2

=4^3 - 3(8)

=64 -24

= 40

integral of sinx dx

= -cosx

evaluated from -pi/2 to pi

= -cospi - (-cos-pi/2)

== -(-1) + 0

= 1