Help with differential equations?

[tennis]

i'm stuck on three problems on my homework - I've been trying to solve them forever! please help? thanks in advance. PS I'm new on here and don't know what to do so I'm just going to type out the problems haha

1) Evaluate: integral of the absolute value of x^2-9 dx on interval [-4, 6]

2) Find F'(x) if F(x)= integral of the sqrt of t^4-2t dt on interval [x, x^3]

3) Evaluate: integral of x/sqrt of 2x-1 dx on interval [1,5]

 

[jhosamelly]

Welcome to PF.

Before anyone can help you with your problem, we need to see what you've started. But ok, I'l give you a hint. for the 1st no... what do you do when you have 2 terms inside the integral? Can you separate them?

 

[tennis]

we haven't really learned this stuff yet in my calc class unfortunately. I've never dealt with absolute values before or intervals with variables..

so i don't really know where to begin or what to do with any of them! just wondering if someone could teach me? and i can do the rest of my homework on my own with these problems as references!

thank youuu

 

[HallsofIvy]

You are doing integrals and have never dealt with absolute values? That is usually dealt with in basic algebra- several years before you start on Calculus.

|x|= x if x\ge 0, |x|= -x if x< 0.

So the question is, where is x^2- 9\ge 0? Where is x^2- 9< 0?
It might help to note that x^2- 9= (x- 3)(x+ 3) and that the product of two factors is positive if and only if the two factors have the same sign. Or you could use the fact that "> 0" and "< 0", for continuous functions such as this, must be separated by "= 0". Where is x^2- 9= 0? The check the sign at one point of each interval.

For the second problem recognize that the problem does NOT ask you to actually do the integral. It does expect you to know "Leibniz formula", an extension of the fundamental theorem of Calculus:
If F(x)= \int_{\alpha(x)}^{\beta(x)} f(x,t) dt then
f(x,\beta(x))\beta&#039;(x)- f(x,\alpha(x))\alpha&#039;(x)+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}(x, t)dt

For
\int\frac{x}{\sqrt{2x- 1}}dx
I recommend the substitution u= 2x- 1. That way du= 2 dx so dx= (1/2)du. Of course, \sqrt{2x-1} becomes \sqrt{u}= u^{1/2}. What is x equal to in terms of u?

You say you "haven't really learned this stuff yet in my calculus class" but obviously your teach thinks you have!

 

[jhosamelly]

Hi tennis,

you need to solve for \left|x^2 -9 \right| = 0 first

of course you'll get x = 3 , -3

then solve for the integral.. since you need to solve the integral on intervals -4 to 6 let's separate this intervals into 3. taking into consideration that x = 3 , -3

solve the integral one at a time

1st would be integrate it with interval from -4 to -3

2nd with interval -3 to 3 (this should be with -(x^2 - 9))

3rd with interval 3 to 6

then add it all up.. that should give you the answer.