Let f(t)=F′(t). Write the integral ∫baf(t)dt and evaluate it using the Fundamental Theorem of Calculus: if F(t)=t4, a=4, and b=7, then the integral is ∫^b_af(t)dt=∫^v2_v1=

Question

where
v1=
v2=
Then, after evaluation,
∫^b_af(t)dt=

Answer 1

Using the Fundamental Theorem of Calculus, we know that

∫baf(t)dt = F(b) - F(a)

Since f(t) = F'(t), we have

f(t) = 4t^3

Therefore,

F(t) = ∫f(t)dt = t^4

Using a=4 and b=7, we have

∫^7_4f(t)dt = F(7) - F(4) = 7^4 - 4^4 = 2401 - 256 = 2145

Therefore,

∫^b_af(t)dt = ∫^7_4f(t)dt = ∫^v2_v14t^3 dt = 2145

To find v1 and v2, we solve the equation

v^4 - 16 = 0

Using the fourth root property, we have

v = ±√2

Since we are looking for the interval [4, 7], we take v1 = -√2 and v2 = √2.

Therefore,

∫^b_af(t)dt = ∫^v2_v14t^3 dt = [(4√2)^4 - (4(-√2))^4]/4 = (512√2 + 512)/4 = 128√2 + 128.

Hence,

∫^b_af(t)dt=128√2 + 128.