Cal's questions at Yahoo Answers regarding the derivative form of the FTOC

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The discussion centers on calculating the derivatives of definite integrals using the Fundamental Theorem of Calculus (FTC). The derivative of the integral F(x) = ∫ from x to 4 of sin(t^4) dt is determined to be F'(x) = -sin(x^4). Similarly, for f(x) = ∫ from x to 12 of t^7 dt, the derivative is f'(x) = -x^7. The application of the FTC and the properties of definite integrals are crucial for deriving these results accurately.

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MarkFL
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Here are the questions:

How to get the derivative of a definite integral?

F(x) = the integral from x to 4 of sin(t^4) dt

F'(x) = ?

----

f(x) = the integral from x to 12 of t^7 dt

f'(x) = ?

I have several more like these. I've been working on them for 6 hours. None of my answers are right. I don't understand what I'm doing wrong!

I have posted a link there to this topic so the OP can see my work.
 
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Hello Cal,

The derivative form of the Fundamental Theorem of Calculus states:

$$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt=f\left(u(x) \right)\frac{du}{dx}$$

Another rule we will need is the following rule for definite integrals:

$$\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx$$

This rule can easily be demonstrated using the anti-derivative form of the Fundamental Theorem of Calculus:

$$\int_a^b f(x)\,dx=F(b)-F(a)=-\left(F(a)-F(b) \right)=-\int_b^a f(x)\,dx$$

So, let's apply these rules to the given problems.

1.) $$F(x)=\int_x^4 \sin\left(t^4 \right)\,dt=-\int_4^x \sin\left(t^4 \right)\,dt$$

Hence:

$$F'(x)=-\frac{d}{dx}\int_4^x \sin\left(t^4 \right)\,dt=-\sin\left(x^4 \right)$$

2.) $$f(x)=\int_x^{12}t^7\,dt=-\int_{12}^x t^7\,dt$$

Hence:

$$f'(x)=-\frac{d}{dx}\int_{12}^x t^7\,dt=-x^7$$
 

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