For the function F(t)=3^t, 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=
(Note that your answer must be exact, not a decimal approximation.)
∫[a,b] f(t) dt = F(b) - F(a) = 3^b - 3^a
To evaluate the integral ∫^{b}_a f(t)dt using the Fundamental Theorem of Calculus, we first need to find the derivative of the function F(t).
Given that F(t) = 3^t, we can use the power rule of differentiation to find f(t), which represents the derivative of F(t). The power rule states that if we have a function of the form f(x) = a^x, where "a" is a constant, then its derivative is given by f'(x) = a^x * ln(a).
In this case, a = 3, so we have f(t) = 3^t * ln(3).
Now, we can write the integral ∫^{b}_a f(t)dt. Given the limits of integration a = 0 and b = 3, the integral becomes:
∫^{3}_0 3^t * ln(3) dt.
Evaluating this integral using the Fundamental Theorem of Calculus involves applying the antiderivative of f(t) and evaluating it at the upper and lower limits of integration.
To find the antiderivative of f(t), we can use the power rule for integration, which states that if we have a function of the form f(x) = a^x, then its antiderivative is given by F(x) = (a^x)/ln(a).
So, the antiderivative of f(t) = 3^t * ln(3) is F(t) = (3^t)/ln(3).
To evaluate the integral ∫^{3}_0 3^t * ln(3) dt, we substitute the upper and lower limits of integration into the antiderivative F(t):
F(3) - F(0) = [(3^3)/ln(3)] - [(3^0)/ln(3)].
Simplifying further:
F(3) - F(0) = [27/ln(3)] - [1/ln(3)].
Finally, the exact value of the integral is [27/ln(3)] - [1/ln(3)], which cannot be simplified further.
To find the integral ∫^{b}_a f(t)dt using the Fundamental Theorem of Calculus, we will first find the function f(t).
Given that F(t) = 3^t, we can find f(t) by taking the derivative of F(t). Using the power rule, the derivative of 3^t is given by:
f(t) = F'(t) = ln(3) * 3^t
Now, we can proceed to evaluate the integral ∫^{b}_a f(t)dt by substituting f(t) into the integral expression:
∫^{3}_0 f(t)dt = ∫^{3}_0 ln(3) * 3^t dt
To evaluate this integral, we can use the antiderivative of f(t), which is F(t) = 3^t. Applying the Fundamental Theorem of Calculus, the evaluated integral is:
F(3) - F(0) = 3^3 - 3^0 = 27 - 1 = 26
Therefore, the value of the integral ∫^{3}_0 f(t)dt is 26.