Derivative of an integral and evaluating an integral

[JennV]

Homework Statement

1.) Evaluate the integral ∫ x*4th root( 25 + x^2)dx

2.) Find dy/dx
y= ∫2pi (upper limit) sin(t^2)dt
sqrt(x) (lower limit)

Homework Equations

Antiderivative

The Attempt at a Solution

1.) I'm not sure how would I find the antiderivative of a product rule derivative. I tried to simplify the derivative which won't make much of a difference because I wouldn't be able to distribute the x into the (25+x^2)^(1/4).

2.) The sqrt(x) is the lower limit and I want it to be the upper limit, so I would times the whole thing by -1. The derivative of an integral function would be the original function by the fundamental theorem of calculus. So it would be -sin((sqrtx)^2) which would be equal -sin(x), but it says that my answer is wrong.

Thank you in advance.

 

[micromass]

For the first one, have you tried the following substitution:

u=\sqrt[4]{25+x^2}

 

[Mark44]

Homework Statement

1.) Evaluate the integral ∫ x*4th root( 25 + x^2)dx

2.) Find dy/dx
y= ∫2pi (upper limit) sin(t^2)dt
sqrt(x) (lower limit)

Homework Equations

Antiderivative

The Attempt at a Solution

1.) I'm not sure how would I find the antiderivative of a product rule derivative. I tried to simplify the derivative which won't make much of a difference because I wouldn't be able to distribute the x into the (25+x^2)^(1/4).

Instead of product rule you should be thinking chain rule. A simple and obvious (to me, at least) substitution will work for this integral.

2.) The sqrt(x) is the lower limit and I want it to be the upper limit, so I would times the whole thing by -1. The derivative of an integral function would be the original function by the fundamental theorem of calculus. So it would be -sin((sqrtx)^2) which would be equal -sin(x), but it says that my answer is wrong.

If you switch the limits of integration the new integral is -1 times the old one, so you're not really multiplying by anything.
You have
y=\int_{\sqrt{x}}^{2\pi} sin(t^2)dt = -\int_{2\pi}^{\sqrt{x}} sin(t^2)dt
\frac{dy}{dx}=-\frac{d}{dx}\int_{2\pi}^{\sqrt{x}} sin(t^2)dt

Now you're almost set, except that the differentation is with respect to x, and the upper limit of integration is not x, but a function of x. Let u = sqrt(x) and use the chain rule. The idea here is that d/dx(f(u)) = d/du(f(u)) * du/dx.

 

[Strants]

For the first one, have you tried the following substitution:

u=\sqrt[4]{25+x^2}

Wouldn't that make du hard to substitute in? I'd think

u=25+x^{2}

would be an easier substitution.

 

[micromass]

No, it's a very easy substitution: if u=\sqrt[4]{25+x^2}, then u^4=25+x^2. Thus 4u^3du=2xdx, which gives us dx=\frac{2u^3}{x}du. The beauty here is, that the x in the denumerator will be eliminated by the x in our integral.

 

[Strants]

Oh, I see. I guess when you take both sides to the fourth, the substitution is pretty easy. I guess I'm just more used to substitutions in the form du = f(x) dx.

 

[Mark44]

For the first one, have you tried the following substitution:

u=\sqrt[4]{25+x^2}

Wouldn't that make du hard to substitute in? I'd think

u=25+x^{2}

would be an easier substitution.

Both substitution work. "There's more than one way to skin a cat." - old saying.

 

[JennV]

Thank you so much to everyone that left a response to my question.
I did manage to get the correct answer for #1, YAY! =)
But I'm still struggling with #2.

Applying what has been said would it be:
Where U is sqrt(x), so du=1/2x^-1/2
So I would put in -sin((1/2x^-1/2)^2), is this the correct answer...?
Sorry I'm asking for confirmation, it's because I only have one attempt left on this question.