Thread: Integral involving e, exponents and fractions

1. Integral involving e, exponents and fractions

Is someone able to show me how I'd calculate this integral (showing working steps)? Any help would be greatly appreciate.

$\int{{-5x^3 e^{-5x}}\over{3}} - {5x^2 e^{-5x} dx}$

2. Originally Posted by drew.walker
Is someone able to show me how I'd calculate this integral (showing working steps)? Any help would be greatly appreciate.

$\int{{-5x^3 e^{-5x}}\over{3}} - {5x^2 e^{-5x} dx}$
This is the same as

$\tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx=\tfrac{1}{3}\int\frac{\,d}{\,dx}\left[x^3e^{-5x}\right]\,dx-6\int x^2e^{-5x}\,dx$ $=\tfrac{1}{3}x^3e^{-5x}-6\int x^2e^{-5x}\,dx$.

Now apply integration by parts twice to $\int x^2e^{-5x}\,dx$ and then you're done!

Can you take it from here?

3. Yeah, I think I'm okay from there. I knew there was a way to simplify the integral, but I just couldn't see it. Been doing this for too long tonight! Thanks.

4. Actually, something I'm confused about. Why did you bother making the equation:

$
\tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx
$

rather than just:
$
\tfrac{1}{3}\int-5x^3e^{-5x}\,dx-\int5x^2e^{-5x}\,dx
$

Wouldn't the second option be equal, but easier to solve?

5. Originally Posted by drew.walker
Actually, something I'm confused about. Why did you bother making the equation:

$
\tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx
$

rather than just:
$
\tfrac{1}{3}\int-5x^3e^{-5x}\,dx-\int5x^2e^{-5x}\,dx
$

Wouldn't the second option be equal, but easier to solve?
Let's see!

Let's focus on $\int -5x^3e^{-5x}\,dx$.

Let $u=-5x^3$ and $\,dv=e^{-5x}\,dx$

Then $\,du=-15x^2\,dx$ and $v=-\tfrac{1}{5}e^{-5x}$

Thus, $\int -5x^3e^{-5x}\,dx=x^3e^{-5x}-3\int x^2e^{-5x}\,dx$.

Now, note that $\tfrac{1}{3}\int -5x^3e^{-5x}\,dx-5\int x^2e^{-5x}\,dx=\tfrac{1}{3}\left[x^3e^{-5x}-3\int x^2e^{-5x}\,dx\right]-5\int x^2e^{-5x}\,dx$ $=\tfrac{1}{3}x^3e^{-5x}-\int x^2e^{-5x}\,dx-5\int x^2e^{-5x}\,dx=\tfrac{1}{3}x^3e^{-5x}-6\int x^2e^{-5x}\,dx$

So you'll end up with the same thing!

I just thought forcing product rule in the integrand would save time in integrating ==> you only need to apply integration by parts twice.