Derivative Formulae for Continuous f(x) on [0,1]

[azatkgz]

Homework Statement

Question 5(10 marks)
Assume that a function f(x) is continuous on the interval [0,1].Express the following derivatives as formulae in terms of
$$x,f(x),f'(x) and \int_{a}^{x}f(t)dt$$

$$a)\frac{d}{dx}\int_{x}^{0}tf(t)dt$$ $$b)\frac{d}{dx}\int_{0}^{x}xf(t)dt$$


$$c)\frac{d}{dx}\int_{0}^{x}xf(x)dt$$

The Attempt at a Solution

I got 6 marks from 10.Can you help me to find my mistakes?

a)$$xf(x)$$


b)$$\frac{d}{dx}\int_{0}^{x}xf(t)dt=\int_{0}^{x}f(t)dt+x\frac{d}{dx}\int_{0}^{x}f<br /> <br /> (t)dt=\int_{0}^{x}f(t)dt+xf(x)$$

c)$$\frac{d}{dx}\int_{0}^{x}xf(x)dt=\int_{0}^{x}f(x)dt+f'(x)x\int_{0}^{x}dt+xf<br /> <br /> (x)\frac{d}{dx}\int_{0}^{x}dt=f(x)x+f'(x)x^2+xf(x)=2xf(x)+x^2f'(x)$$

 

[quasar987]

c) is right

for a), the FTC says that

$$\frac{d}{dx}\int_{0}^{x}tf(t)dt=xf(x)$$

so you forgot to invert the bounds of the integrals first, which would induce a minus sign in the answer.

For b), this is an application of Leibniz rule of differentiation under the integral sign where the integrand xf(t) is considered as a function of two variables g(x,t)=xf(t).

 

[azatkgz]

then $$\frac{d}{dx}\int_{0}^{x}g(x,t)dt$$.


Right?

 

[quasar987]

what's your question?

 

[azatkgz]

$$\frac{d}{dx}\int_{0}^{x}g(x,t)dt=\frac{d}{dx}\int_{0}^{x}xf(x)dt$$

 

[quasar987]

No,

$$\frac{d}{dx}\int_{0}^{b(x)}g(x,t)dt = \int_{0}^{b(x)}\frac{\partial}{\partial x}g(x,t)dt+b'(x)g(x,b(x))$$

So

$$\frac{d}{dx}\int_{0}^{x}xf(t)dt = \int_{0}^{x}f(t)dt+xf(x)$$

 

[azatkgz]

Very nice.I've haven't seen this method.