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.