# Thread: Confusing integrals

1. ## Confusing integrals

These integrals don't make any sense. I know the basics but these are out of my legaue. I don't even know how to write the integral for that it has a top number and a bottom number to it. So ill give it a shot :

1) ∫^1 and the bottom is 0 then (4x^3 - 2x)dx

2) ∫^pie(3.14) and the bottom is pie/2 then (sin2x dx)

3) ∫^2 and the bottom is 0 then [3dx/ (4x +1)^1/2]

I know its confusing to read but i really need help on this to be ready for my quiz. So thanks for the help.

2. I will do the first one as an example:
Originally Posted by uniquereason81
1) ∫^1 and the bottom is 0 then (4x^3 - 2x)dx
Initial equation:
$\int_{0}^{1} (4x^3 - 2x) ~dx$

Integrate:
$=x^4 - x^2 |_0^1$

The line on the right tells you to evaluate it from 0 to 1.
This means that if f(x) is our function: $f(x) = x^4 -x^2$
Then we are looking for f(1)-f(0)

So:

$=[(1)^4 -(1)^2] - [(0)^4 - (0)^2]$

$=[1 -1] - [0 - 0]$

$=0$

3. Thanks but my teacher was saying we could sub x1 and x2 as u1 and u2 by finding it in the equation. Or something like that, is that correct

4. ## For the third one

Originally Posted by uniquereason81
These integrals don't make any sense. I know the basics but these are out of my legaue. I don't even know how to write the integral for that it has a top number and a bottom number to it. So ill give it a shot :

1) ∫^1 and the bottom is 0 then (4x^3 - 2x)dx

2) ∫^pie(3.14) and the bottom is pie/2 then (sin2x dx)

3) ∫^2 and the bottom is 0 then [3dx/ (4x +1)^1/2]

I know its confusing to read but i really need help on this to be ready for my quiz. So thanks for the help.
you have $\int_0^{2}\frac{3}{\sqrt{4x+1}}dx\Rightarrow{\frac {3}{4}\int_0^{2}\frac{4}{\sqrt{4x+1}}}dx$...now you should see that there is a quantity and a derivative of that quantity

5. Though how would you start number 3. Thats the most confusing one for me to do and also start

6. ## Uhm

Originally Posted by uniquereason81
Though how would you start number 3. Thats the most confusing one for me to do and also start
Didnt I just give you a hint? I am sorry did I do the wrong one?

7. ## Ok

Originally Posted by uniquereason81
Though how would you start number 3. Thats the most confusing one for me to do and also start
Well you are not responding so I will just show you we got this far $\frac{3}{4}\int_0^{2}\frac{4}{\sqrt{4x+1}}dx$...now that we have an integral of the form $\int{f(g(x))\cdot{g'(x)}dx}$ (since the derivative of [tex]4x+1{/math] is 4) then we just integrate it as though it was just f(x)...so we would get $\int_0^{2}\frac{3}{\sqrt{4x+1}}dx=\bigg[\frac{3}{4}\cdot{2}\sqrt{4x+1}\bigg]\bigg|_0^{2}=3$

8. ooo my bad i didn't see your post.

9. ## Haha

Originally Posted by uniquereason81
ooo my bad i didn't see your post.
Don't worry about it

10. How would you start number 2

11. ## Ok

Originally Posted by uniquereason81
How would you start number 2
Same concept you have $\int_{\frac{\pi}{2}}^{\pi}sin(2x)dx$...to integrate that we need the derivative of the inside of the sine function to be on the outside...the derivative is 2...but to put a two we need to counter with a half...so we would have $\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}2sin(2x)dx\Ri ghtarrow{\frac{-cos(2x)}{2}\bigg|_\frac{\pi}{2}^{\pi}}=-1$

12. Originally Posted by uniquereason81
How would you start number 2
Follow my example.

Integrate the function exactly as you normally would for an indefinite integral (one with no values at the bottom and top). After it is integrated, evaluate it like this:

integral from a to b of f'(x) = f(x) from a to b = f(b) - f(a)

You can copy my example almost exactly.