Given the definite integral F(x)=∫1220x(h4+63hh5)dhF\left(x\right)=\int_{12}^{20x}\!\left(h^4+\frac{63h}{\sqrt{h^5}}\right)\,dhF(x)=∫1220x(h4+h563h)dh, find the derivative F′(x)F^{\prime}\left(x\right)F′(x).

F ′ ( x ) = 20 5 x 4 + 25 , 200 x ( 20 x ) 5 F^{\prime}\left(x\right)=20^5x^4+\frac{25,200x}{\sqrt{\left(20x\right)^5}}

Given the definite integral F ( x ) = ∫ 12 20 x ( h 4 + 63 h h 5 ) d h F\left(x\right)=\int_{12}^{20x}\!\left(h^4+\frac{63h}{\sqrt{h^5}}\right)\,dh F ( x ) = ∫ 12 20 x ( h 4 + h 5 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> 63 h ) d h , find the derivative F ′ ( x ) F^{\prime}\left(x\right) F ′ ( x ) .

So in this problem, we are given that big f of x is equal to this definite integral right here, and we're asked to find the derivative of this integral, big f prime of x. So to solve this problem, what we're doing is we're taking the derivative of an integral. So notice that we're taking the derivative of this function, which would be the derivative with respect to x. It's with respect to x here of this integral. It's gonna go from 12 to 20 x, and it's gonna be multiplied by this quantity right here. So we have h to the fourth power plus 63 h over the square root of h to the fifth. And then that whole thing is with respect to h. So this is what we end up getting. Now from here, what I need to do is think about what this is all gonna come out to be. And I can use the fundamental theorem of calculus to do that. Because notice that we are taking the derivative of an integral, which means we can apply the fundamental theorem. Now my first rule of the fundamental theorem says I can take whatever this high bound is and plug it in for h if we have a function of x up here, which we do, it's 20 x. So let's go ahead and do that. So this derivative is going to come out to, and in this case we'll take h and we'll replace it with 20 x. So that's gonna be 20 x to the fourth power. We're gonna have plus, and it's gonna be 63 times 20 x. And all that's going to be divided by the square root of 20 x to the fifth power. Now we're not quite done here yet, because notice that we have a function of x rather than just x. And whenever this happens, what you need to do is use the chain rule. And the way you can use the chain rule and in combination with the fundamental theorem is to take whatever this high bound is and take its derivative and multiply it by the outside of the whole quantity you calculated. The derivative of 20 x with respect to x would just give you 20. So we multiply 20 for this high bound here, and then that's going to be our solution, but we can go ahead and simplify this a little bit more. So if we wanna simplify things, well, I can take this 20 and I can bring it out to the front here. I can multiply it by this 20 x to the fourth, and that's going to give me 20 They're going to have to the fifth power times x to the fourth power. The reason for this is I could take that fourth power that I could put on the 20, and then multiplying it by this could be 20 to the fifth, and then I could put it into that x to the fourth to give me that right there. Then I'm going to have plus, and they're going to have 20 times 63. That's gonna be multiplied by 20 x. I'm just gonna write it like this. And all of that's going to be divided by the square root of 20 x to the fifth power. So this is how we can expand things out. And then from here, I'm just gonna multiply these numbers on a calculator just to make it quick. And that's going to give me 20 to the fifth power times x to the fourth power right up front, and then we're gonna have plus. And then this whole thing comes up to 25,200. It's gonna be multiplied by x all divided by this quantity out here, the square root of 20 x to the fifth power. So this is how you can solve this problem. We just do a little bit of work at the end to simplify this. But as you can see, using the fundamental theorem allows us to give this in a much more quick way than trying to find what this integral is and then take its derivative and go through that long process. Instead, we're able to just substitute in the top down for the h, and then we were just needed to multiply this by the derivative, and that gave us our solution. So that is how you can use the fundamental theorem of calculus part one. Let's move on to the next one.

Given the definite integral F ( x ) = ∫ 12 20 x ⁣ ( h 4 + 63 h h 5 ) d h F\left(x\right)=\int_{12}^{20x}\!\left(h^4+\frac{63h}{\sqrt{h^5}}\right)\,dh F ( x ) = ∫ 12 20 x ( h 4 + h 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 63 h ) d h , find the derivative F ′ ( x ) F^{\prime}\left(x\right) F ′ ( x ) .

F ′ ( x ) = 20 5 x 4 + 25 , 200 x ( 20 x ) 5 F^{\prime}\left(x\right)=20^5x^4+\frac{25,200x}{\sqrt{\left(20x\right)^5}}

F ′ ( x ) = ( 20 x ) 4 + 1260 x ( 20 x ) 5 F^{\prime}\left(x)=\left(20^{}x\right.\right)^4+\frac{1260x}{\sqrt{\left(20x\right)^5}}

F ′ ( x ) = ( 20 x ) 5 + 25 , 200 x 2 ( 20 x ) 5 F^{\prime}\left(x\right)=\left(20x\right)^5+\frac{25,200x^2}{\sqrt{\left(20x\right)^5}}

F ′ ( x ) = ( 20 x ) 5 + 1260 x 2 ( 20 x ) 5 F^{\prime}\left(x\right)=\left(20x\right)^5+\frac{1260x^2}{\sqrt{\left(20x\right)^5}}

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 $F\left(x\right)=\int_{12}^{20x}\!\left(h^4+\frac{63h}{\sqrt{h^5}}\right)\,dh$ , find the derivative $F^{\prime}\left(x\right)$ .

$F^{'} (x) = 20^{5} x^{4} + \frac{25, 200 x}{\sqrt{{(20 x)}^{5}}}$

$F^{'} {(x) = (20^{} x)}^{4} + \frac{1260 x}{\sqrt{{(20 x)}^{5}}}$

$F^{'} (x) = {(20 x)}^{5} + \frac{25, 200 x^{2}}{\sqrt{{(20 x)}^{5}}}$

$F^{'} (x) = {(20 x)}^{5} + \frac{1260 x^{2}}{\sqrt{{(20 x)}^{5}}}$

Verified step by step guidance

Recognize that the problem involves finding the derivative of a definite integral with a variable upper limit, which suggests the use of the Fundamental Theorem of Calculus Part 1.

According to the Fundamental Theorem of Calculus Part 1, if F(x) = ∫[a, g(x)] f(t) dt, then F'(x) = f(g(x)) * g'(x). Here, a = 12, g(x) = 20x, and f(h) = h^4 + 63h / sqrt(h^5).

Substitute g(x) = 20x into the function f(h) to get f(20x) = (20x)^4 + 63(20x) / sqrt((20x)^5).

Differentiate g(x) = 20x with respect to x to find g'(x). Since g(x) = 20x, g'(x) = 20.

Apply the chain rule: F'(x) = f(g(x)) * g'(x) = [(20x)^4 + 63(20x) / sqrt((20x)^5)] * 20.

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