Partial Derivative of Integral

[Victor8108]

Homework Statement

Find df/dx, f(x,y)=integral of sqrt(1-t^3)dt from x^2 to x^3.

Since it is asking to find the derivative with respect to x,should I regard t as a constant?

Homework Equations

The Attempt at a Solution

I tried to find the antiderivative of the integral sqrt(1-t^3)dt and then differentiate with respect to x.

However, I'm stuck in finding the antiderivative of the function , I tried using substitution u=1-t^3, du=-3t^2 dt. But that doesn't seem to be going the right direction.

 

[slider142]

The variable "t" is a dummy variable we are summing over: the value of the definite integral is not a function of t. To see this another way, apply the Fundamental Theorem of Calculus to write the integral as a difference of two functions of x. You do not actually need to find an expression for these two functions, as all you will need are their derivatives. For example, consider
$$f(x) = \int_0^{x^2} \sin(t^2) dt$$
The Fundamental Theorem of Calculus tells us that we can write f(x) like this as well, for each value of x:
$$f(x) = F(x^2) - F(0)$$
where F(t) is any function whose derivative is $\sin(t^2)$ at each point t in the closed interval between t = 0 and t = $x^2$. If we try to take the derivative of f(x), we find the following when we apply the chain rule:
$$f'(x) = F'(x^2)(2x) - F'(0)(0)$$
We don't need to know any particular version of F, we just need to know what its derivative at $x^2$ and 0 is. Since $x^2$ and 0 are in the closed interval, we know the value of the derivative of F at those points are $\sin(x^4)$ and sin(0), respectively. Thus we get
$$f'(x) = 2x\sin(x^4)$$
once we simplify.

 

[Victor8108]

For those wondering why the equation f(x,y) has no y, here is the complete function:

f(x,y)= (x^2)y+integral of sqrt(1-t^3)dt from x^2 to x^3.

I figured the first part(x^2)y so I didnt put it in the equation. sorry about the inconvenience.

 

[Victor8108]

Ok, according to the fundamental theorem of calculus:

if g(x)= f(t)dt from a to x, then g'(x)=f(x)

applying this to my problem, then it becomes g'(x)= sqrt(1-x^9)-sqrt(1-x^6)

 

[voko]

The fundamental theorem states that if F(a, b) is the integral of f(x) from a to b, then, letting g(t) = F(a, t), g'(t) = f(t).

 

[slider142]

Ok, according to the fundamental theorem of calculus:

if g(x)= f(t)dt from a to x, then g'(x)=f(x)

applying this to my problem, then it becomes g'(x)= sqrt(1-x^9)-sqrt(1-x^6)

That's absolutely right.