integrals help

**break** · December 6th 2009, 01:39 PM

Compute an antiderivative for each of the following functions:

$a) f(x) = \frac{2x^3+1}{x^2}$
$b) g(x) = \sqrt{x}+x^\frac{7}{3}$
$c) h(x) = csc(2x)cot(2x)$

I was never taught integrals properly, only the basics where u change $x^2$ to $x^3/3$ .

can anyone help? thanks!

**adkinsjr** · December 6th 2009, 02:02 PM

Originally Posted by break

Compute an antiderivative for each of the following functions:

$a) f(x) = \frac{2x^3+1}{x^2}$
$b) g(x) = \sqrt{x}+x^\frac{7}{3}$
$c) h(x) = csc(2x)cot(2x)$

I was never taught integrals properly, only the basics where u change $x^2$ to $x^3/3$ .

can anyone help? thanks!

For a polynomial $ax^n$ the integral is given by $a\int x^ndx=\frac{a}{n+1}x^{n+1}+C$ . Convince yourself this is true by differentiating.

The first one can be written as a sum:

$\int\frac{2x^3+1}{x^2}dx=\int2xdx+\int\frac{1}{x^2 }dx$

Use the formula. Remember that $\frac{1}{x^2}=x^{-2}$ .

Problem (b) just requires the formula. Remember that $\sqrt{x}=x^{\frac{1}{2}}$ .

For (c) rememeber that $\frac{d}{du}csc(u)=-csc(u)cot(u)$

So if you let $u=2x$ so that $du=2dx$ then $csc(2x)cot(2x)dx=\frac{1}{2}csc(u)cot(u)du$ . Does this make sense? Do you see what the integral is?

**adkinsjr** · December 6th 2009, 02:07 PM

I had to edit my original post for latex error. Sorry about that.

**break** · December 6th 2009, 02:24 PM

Originally Posted by adkinsjr

The first one can be written as a sum:

$\int\frac{2x^3+1}{x^2}dx=\int2xdx+\int\frac{1}{x^2 }dx$

Use the formula. Remember that $\frac{1}{x^2}=x^{-2}$ .

ok so this turns to

$\int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

$\int x^2dx + \int\frac{-1}{x}dx$

i think i did something wrong

**Prove It** · December 6th 2009, 02:30 PM

Originally Posted by break

Compute an antiderivative for each of the following functions:

$a) f(x) = \frac{2x^3+1}{x^2}$
$b) g(x) = \sqrt{x}+x^\frac{7}{3}$
$c) h(x) = csc(2x)cot(2x)$

I was never taught integrals properly, only the basics where u change $x^2$ to $x^3/3$ .

can anyone help? thanks!

Alternatively, for c), remember

$\csc{(2x)}\cot{(2x)} = \frac{1}{\sin{(2x)}}\cdot \frac{\cos{(2x)}}{\sin{(2x)}}$

$= \frac{\cos{(2x)}}{\sin^2{(2x)}}$ .

Now use a $u$ substitution.

Let $u = \sin{(2x)}$ so that $\frac{du}{dx} = 2\cos{(2x)}$ .

Then you have

$\int{\csc{(2x)}\cot{(2x)}\,dx} = \int{\frac{\cos{(2x)}}{\sin^2{(2x)}}\,dx}$

$= \frac{1}{2}\int{\frac{1}{\sin^2{(2x)}}[2\cos{(2x)}]\,dx}$

$= \frac{1}{2}\int{\frac{1}{u^2}\,\frac{du}{dx}\,dx}$

$= \frac{1}{2}\int{u^{-2}\,du}$

$= -\frac{1}{2}u^{-1} + C$

$= -\frac{1}{2}\cdot\frac{1}{\sin{(2x)}} + C$

$= -\frac{1}{2}\csc{(2x)} + C$ .

**break** · December 6th 2009, 02:36 PM

Originally Posted by Prove It

Then you have

$\int{\csc{(2x)}\cot{(2x)}\,dx} = \int{\frac{\cos{(2x)}}{\sin^2{(2x)}}\,dx}$

$= -\frac{1}{2}\int{\frac{1}{\sin^2{(2x)}}[-2\cos{(2x)}]\,dx}$

i got lost here. where did u get the -1/2 from? is there a rule for this?

**adkinsjr** · December 6th 2009, 02:36 PM

Originally Posted by break

ok so this turns to

$\int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

$\int x^2dx + \int\frac{-1}{x}dx$

i think i did something wrong

You're way off. The integral $\int 2xdx=\frac{2}{3}x^3+C$ is given by the formula $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$ . Do you see how I applied the formula? It works the same for all polynomials. See if you can do the same for the second term.

Here is a tutorial that might help you with some of these. It seems like you don't have the basic concept of what an integral or anti-derivative is.

Pauls Online Notes : Calculus I - Indefinite Integrals

**Prove It** · December 6th 2009, 02:39 PM

Originally Posted by break

i got lost here. where did u get the -1/2 from? is there a rule for this?

1. Notice my edit.

2. We need to change the integrand so that it has $\frac{du}{dx}$ in it.

So we change $\cos{(2x)}$ into $\frac{1}{2}\cdot 2\cos{(2x)}$ .

Then I took the $\frac{1}{2}$ outside the integral.

**break** · December 6th 2009, 02:53 PM

Originally Posted by adkinsjr

You're way off. The integral $\int 2xdx=\frac{2}{3}x^3+C$ is given by the formula $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$ . Do you see how I applied the formula? It works the same for all polynomials. See if you can do the same for the second term.

Here is a tutorial that might help you with some of these. It seems like you don't have the basic concept of what an integral or anti-derivative is.

Pauls Online Notes : Calculus I - Indefinite Integrals

why does x skip over $x^2$ and go to $x^3$ ? I see how the formula says n+1 but, 1+1 = 3? i don't get it.

**adkinsjr** · December 6th 2009, 02:56 PM

Originally Posted by break

why does x skip over $x^2$ and go to $x^3$ ? I see how the formula says n+1 but, 1+1 = 3? i don't get it.

You're right

It should be $x^2+C$ .

$\int 2xdx =\frac{2}{1+1}x^{1+1}+C=x^2+C$

Sorry about that. That was just a stupid mistake on my part.

**break** · December 6th 2009, 03:00 PM

Originally Posted by adkinsjr

You're right

It should be $x^2+C$ .

$\int 2xdx =\frac{2}{1+1}x^{1+1}+C=x^2+C$

Sorry about that. That was just a stupid mistake on my part.

haha my brain was seriously about to explode right now. phew.

Originally Posted by break

$\int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

$\int x^2dx + \int\frac{-1}{x}dx$

i think i did something wrong

so i did get the first part right... right? haha. do i still have to plug in a + C early on?

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