Let f(t)=F′(t). Write the integral ∫^{b}_a f(t)dt and evaluate it using the Fundamental Theorem of Calculus: if F(t)=t^5, a=1, and b=4, then the integral is :
∫^{b}_a f(t)dt=∫^{v2}_v1 =
where
v1=
v2=
Then, after evaluation,
∫^{b}_a f(t)dt=
c'mon, these are all the same.
∫[1,4] f(t) = F(4)-F(1) = 4^5 - 1^5 = 1023