1. ## evaluate

Given that $y=2xe^{2x-1}$ and $\frac{dy}{dx}=(4x+2)(e^{2x-1})$, hence evaluate $\int^1_0[2xe^{2x-1}]dx$

2. by parts dv = exp(2x-1) u=x

$\displaystyle \int 2x e^{2x-1} \; dx =\frac{2x e^{2x-1}}{2} - 2\int e^{2x-1} dx = \frac{2xe^{2x-1}}{2} - \frac{2e^{2x-1}}{2}+C$

$\displaystyle \int_{0}^{1} 2xe^{2x-1} dx = e^{2-1} - e^{2-1} - ( 0 - e^{0 -1 }) = \frac{1}{e}$

3. answer is 1.54, $\frac{1}{e}=3.67$ which is wrong

4. Use integration by parts:

Let $u = 2x$ and $dv=e^{2x-1}$

so, $du = 2dx$ and $v = \frac{e^{2x-1}}{2}$

Now,

$\int 2x e^{2x-1} dx = 2x.\frac{1}{2} e^{2x-1} - \int \frac{1}{2}e^{2x-1}2dx$

$= x(e^{2x-1}) - \frac{e^{2x-1}}{2}$

$\int_{0}^{1} 2xe^{2x-1} dx = \left[ e - \frac{e}{2} \right] - \left[0 - \frac{1}{2e}\right] = \frac{e}{2} + \frac{1}{2e} = 1.543$

5. Originally Posted by Punch
Given that $y=2xe^{2x-1}$ and $\frac{dy}{dx}=(4x+2)(e^{2x-1})$, hence evaluate $\int^1_0[2xe^{2x-1}]dx$
Don't use integration by parts, you're asked to use the information given.

$\displaystyle{\frac{dy}{dx} = (4x + 2)e^{2x - 1}}$

$\displaystyle{\frac{dy}{dx} = 4x\,e^{2x - 1} + 2e^{2x - 1}}$

You should know $\displaystyle{y = \int{\frac{dy}{dx}\,dx}}$, so

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1} + 2e^{2x - 1}\,dx}}$

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} + \int{2e^{2x - 1}\,dx}}$

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x- 1}\,dx} + e^{2x - 1}}$

$\displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx}}$

$\displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = 2\int{2x\,e^{2x - 1}\,dx}}$

$\displaystyle{x\,e^{2x - 1} - \frac{e^{2x - 1}}{2} = \int{2x\,e^{2x - 1}\,dx}}$.

Now evaluate $\displaystyle{\int_0^1{2x\,e^{2x - 1}\,dx}}$.

6. Originally Posted by Amer
by parts dv = exp(2x-1) u=x

$\displaystyle \int 2x e^{2x-1} \; dx =\frac{2x e^{2x-1}}{2} - 2\int e^{2x-1} dx = \frac{2xe^{2x-1}}{2} - \frac{2e^{2x-1}}{2}+C$

$\displaystyle \int_{0}^{1} 2xe^{2x-1} dx = e^{2-1} - e^{2-1} - ( 0 - e^{0 -1 }) = \frac{1}{e}$
The answer came out to be wrong here because Amer supposed dv = exp(2x-1),, which gives $v = \frac{exp(2x-1)}{2}$

and applying integration by parts would give:

$uv - \int v. du$

$\displaystyle \int 2x e^{2x-1} \; dx =\frac{2x e^{2x-1}}{2} - 2\int \frac{e^{2x-1}}{2} dx = \frac{2xe^{2x-1}}{2} - \frac{2e^{2x-1}}{4}+C$

7. Originally Posted by Prove It
Don't use integration by parts, you're asked to use the information given.

$\displaystyle{\frac{dy}{dx} = (4x + 2)e^{2x - 1}}$

$\displaystyle{\frac{dy}{dx} = 4x\,e^{2x - 1} + 2e^{2x - 1}}$

You should know $\displaystyle{y = \int{\frac{dy}{dx}\,dx}}$, so

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1} + 2e^{2x - 1}\,dx}}$

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} + \int{2e^{2x - 1}\,dx}}$

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x- 1}\,dx} + e^{2x - 1}}$

$\displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx}}$

$\displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = 2\int{2x\,e^{2x - 1}\,dx}}$

$\displaystyle{x\,e^{2x - 1} - \frac{e^{2x - 1}}{2} = \int{2x\,e^{2x - 1}\,dx}}$.

Now evaluate $\displaystyle{\int_0^1{2x\,e^{2x - 1}\,dx}}$.
after evaluating, ans is 5.436 which is still wrong

Also, I did understand this part!!

from $\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x- 1}\,dx} + e^{2x - 1}}$

to $\displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx}}$

Lastly, indeed i am asked to use the information given!! pls do not use integration by part thanks!

8. Prove It has already shown that :

$\int{2x\,e^{2x - 1}\,dx}} = \displaystyle{x\,e^{2x - 1} - \frac{e^{2x - 1}}{2}$

So,

$\displaystyle{\int_0^1{2x\,e^{2x - 1}\,dx}} = \left[ x e^{2x-1} - \frac{e^{2x-1}}{2} \right]_0^1$ gives you 1.543

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} + \int{2e^{2x - 1}\,dx}}$

and,

$\int 2e^{2x - 1} dx= 2 \int e^{2x - 1}dx = 2 \times \frac{e^{2x-1}}{2} = e^{2x-1}$

So,

$\displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} + e^{2x-1}$

then, $e^{2x-1}$ is subtracted form both the sides to get:

$\displaystyle{2x\,e^{2x - 1} - e^{2x-1} = \int{4x\,e^{2x - 1}\,dx}$

9. Great!! i love MFH