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=
where
v1=
v2=
Then, after evaluation,
∫^b_af(t)dt=
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.
To evaluate the integral ∫ba f(t) dt, we will use the Fundamental Theorem of Calculus.
Given that F(t) = t^4, and a = 4, and b = 7, we can find f(t) by taking the derivative of F(t).
f(t) = F'(t) = d/dt (t^4) = 4t^3
Now, let's evaluate the integral using the given values:
∫ba f(t) dt = ∫^7_4 f(t) dt
Plugging in f(t) = 4t^3:
∫^7_4 4t^3 dt
To integrate this expression, we apply the power rule for integration.
∫^7_4 4t^3 dt = [t^4]_4^7
Evaluating at the upper limit (7) and the lower limit (4):
[t^4]_4^7 = 7^4 - 4^4
Simplifying the expression:
7^4 = 2401
4^4 = 256
2401 - 256 = 2145
Therefore, ∫^7_4 4t^3 dt = 2145.
To evaluate the integral ∫baf(t)dt using the Fundamental Theorem of Calculus, we need to first find F(t) by integrating f(t). In this case, we are given that F'(t) = f(t), and we need to find F(t).
Since f(t) = F'(t), we can find F(t) by integrating f(t) with respect to t. Let's integrate the term t^4 with respect to t to find F(t):
∫t^4 dt = (t^5)/5 + C
Here, C represents the constant of integration.
Now that we have found F(t), we can evaluate the integral ∫baf(t)dt using the Fundamental Theorem of Calculus. According to the theorem, the integral of a function f(t) from a to b is equal to the antiderivative evaluated at b minus the antiderivative evaluated at a.
In this case, we have a = 4 and b = 7. Let's substitute these values into the antiderivative we found:
F(7) = (7^5)/5 + C
F(4) = (4^5)/5 + C
To evaluate the integral, we subtract the antiderivative at a from the antiderivative at b:
∫^b_af(t)dt = [F(t)]^b_a = (7^5)/5 + C - (4^5)/5 - C
Since the constant C appears on both sides of the equation, it cancels out:
∫^b_af(t)dt = (7^5)/5 - (4^5)/5
Therefore, ∫^b_af(t)dt = (16807)/5 - (1024)/5 = 3361.4 - 204.8 = 3156.6.