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 $\int_0^{2}\frac{3}{\sqrt{4x+1}}dx=\bigg[\frac{3}{4}\cdot{2}\sqrt{4x+1}\bigg]\bigg|_0^{2}=3$

9. ## Haha

Originally Posted by uniquereason81

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