For the function F(t)=et2
, let f(t)=F′(t). Write the integral ∫^b_a f(t)dt and evaluate it with the Fundamental Theorem of Calculus.
∫^3_0=
dt=
f(t) = 2tet2
Using the Fundamental Theorem of Calculus, we have:
∫^b_a f(t)dt = F(b) - F(a)
Plugging in the values of a and b, we get:
∫^3_0 f(t)dt = F(3) - F(0)
F(t) = et2
F(3) = e(3)2
F(0) = e(0)2 = 1
∫^3_0 f(t)dt = F(3) - F(0) = e(3)2 - 1 ≈ 19.0857.