Given the definite integral F(x)=∫3x[t8−sin(t4)]dtF\left(x\right)=\int_3^{x}\!\left\lbrack t^8-\sin\left(t^4\right)\right\rbrack\,dt , find the derivative F′(x)F^{\prime}\left(x\right).

F ′ ( x ) = x 8 − sin ⁡ ( x 4 ) F^{\prime}\left(x\right)=x^8-\sin\left(x^4\right)

Given the definite integral F ( x ) = ∫ 3 x [ t 8 − sin ( t 4 ) ] d t F\left(x\right)=\int_3^{x}\!\left\lbrack t^8-\sin\left(t^4\right)\right\rbrack\,dt , find the derivative F ′ ( x ) F^{\prime}\left(x\right) .

So in this problem, we are given the definite integral big f of x, which is this function right here, and we're asked to find the derivative of this function f prime of x. So what we're trying to do is find f prime, which is gonna be the derivative of big f of x. So that means what we have is this whole integral set to big f of x. So all I'm doing is taking the derivative of this entire integral. So it's gonna be the integral from three to x, and it's going to be of t to the eighth power minus the sine of t to the fourth power. Now notice in this situation that we're taking the derivative of an integral. We can apply the fundamental theorem of calculus to solve this problem, the first part of it. And the way that we can do that is by remembering since we just have an x in the upper bound of this this function, what I can do is just take the x and replace it with all the t's that I see. So this whole thing is just gonna come out to we have t to the eighth, so it's gonna be x to the eighth. We're going to have minus, then we have sine t to the fourth, so that's gonna be sine x to the fourth power within this function right here. And then that's going to be the solution to the problem, and that is how you can solve this. So as you can see, using the fundamental theorem of calculus, solving these types of problems becomes very straightforward. And remember, this only works if you have just an x in the top bound of your function. If you had a function of x, you need to use the chain rule. But since we just had an x, we could get away with just taking all these ts and replacing them with xs. So hope you found this video helpful, and let's move on.

Given the definite integral F ( x ) = ∫ 3 x ⁣ [ t 8 − sin ⁡ ( t 4 ) ] d t F\left(x\right)=\int_3^{x}\!\left\lbrack t^8-\sin\left(t^4\right)\right\rbrack\,dt , find the derivative F ′ ( x ) F^{\prime}\left(x\right) .

F ′ ( x ) = x 8 − sin ⁡ ( x 4 ) F^{\prime}\left(x\right)=x^8-\sin\left(x^4\right)

F ′ ( x ) = 3 x 9 − 3 x sin ⁡ ( x 4 ) F^{\prime}\left(x\right)=3x^9-3x\sin\left(x^4\right)

F ′ ( x ) = x 9 − x sin ⁡ ( x 4 ) F^{\prime}\left(x\right)=x^9-x\sin\left(x^4\right)

F ′ ( x ) = 3 x 8 − 3 sin ⁡ ( x 4 ) F^{\prime}\left(x\right)=3x^8-3\sin\left(x^4\right)

8. Definite Integrals

Fundamental Theorem of Calculus

Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

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Multiple Choice

Given the definite integral $F (x) = \int_{3}^{x} [t^{8} - \sin (t^{4})] d t$ , find the derivative $F^{'} (x)$ .

$F^{'} (x) = 3 x^{9} - 3 x \sin (x^{4})$

$F^{'} (x) = x^{9} - x \sin (x^{4})$

$F^{'} (x) = 3 x^{8} - 3 \sin (x^{4})$

$F^{'} (x) = x^{8} - \sin (x^{4})$

Verified step by step guidance

Recognize that the problem involves finding the derivative of a definite integral with a variable upper limit, which is a classic application of the Fundamental Theorem of Calculus.

According to the Fundamental Theorem of Calculus, if F(x) = ∫[a, x] f(t) dt, then the derivative F'(x) = f(x).

Identify the integrand function f(t) = t^8 - sin(t^4) from the given integral F(x) = ∫[3, x] (t^8 - sin(t^4)) dt.

Apply the Fundamental Theorem of Calculus: F'(x) = f(x) = x^8 - sin(x^4).

Verify that the derivative F'(x) = x^8 - sin(x^4) matches the correct answer provided in the problem statement.

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