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Find the derivative using the Fundamental Theorem of Calculus, part 1, which states that if f(x) is continuous over an interval [a, b], and the function F(X) is defined by F(X) - F'(x) = f(x) over [a, b]. [*re) or, then del tot Additional Materials eBook 4.
then Find the derivative using the Fundamental Theorem of Calculus, part 1, which states that if f(x) is continuous over an interval (a, b), and the function F(x) is defined by F(X) - F'(x) = f(x) over [a, b]. de la intu In(u) du 6.
Evaluate the definite integral using the Fundamental Theorem of Calculus, part 2, which states that iffis continuous over the interval [a, b] and F(x) is any antiderivative of f(x), then Giºrno dx = F(b) – Fla). [x?- ») 3x) dx Х

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On solving the question, by the help of, Fundamental Theorem of Calculus, we got answer  [tex]3xe^x[/tex]

What is fundamental Theorem of Calculus?

A theorem that connects the ideas of differentiating a function from integrating a function is known as the fundamental theorem of calculus. The two processes are inverses of one another, with the exception of a constant number that is dependent on the starting point for computing area.

When asked to compute the derivative of an integral, utilize the calculus fundamental theorem rather than evaluating the integral.

FTOC informs us that:

[tex]\frac{d}{dx} [\int\limits^{3x}_t {3lnt} \, dxt] = 3xe^x[/tex]

(That is, the integral's derivative returns the original function to us.)

We are required to locate:

h'(x) where h(x)= [tex]\int\limits^{3x}_t {3lnt} \, dxt= 3xe^x[/tex]

what we want

[tex]h'(x) = \frac{d}{dx} [\int\limits^{ex}_t {3lnt} \, dxt] = 3xe^x.............................A[/tex]

(Note that the first integral's upper limits are not in the proper format for the FTOC to be applied immediately.) The chain rule and a replacement can be used to alter the definite integral. Let:

[tex]u = e^x = \frac{du}{dx} = e^x[/tex]

Using the chain rule and substituting into the integral [A], we obtain:

[tex]\frac{d}{dx} [\int\limits^{ex}_t {3lnt} \, dxt] =[/tex][tex]\frac{d}{dx} [\int\limits^{u}_t {3lnt} \, dxt] = 3xe^x[/tex]

= [tex]e^x \frac{d}{du} [\int\limits^{ex}_t {3lnt} \, dxt] = 3xe^x[/tex]

now we got answer - [tex]3xe^x[/tex]

To know more about Fundamental Theorem of Calculus visit:

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