Integral of t^3*(e^4t^2) and t^2*(e^4t)

**paulbk108** · December 6th 2011, 12:42 AM

Been doing plenty of practice of late just to get back in the swing of things. Could someone please help me out with the solutions to the above?

I think they may involve integration by parts twice, which sounds ugly.

I can't find any similar examples in notes or text book...

Regards.

**princeps** · December 6th 2011, 01:17 AM

Originally Posted by paulbk108

Been doing plenty of practice of late just to get back in the swing of things. Could someone please help me out with the solutions to the above?

I think they may involve integration by parts twice, which sounds ugly.

I can't find any similar examples in notes or text book...

Regards.

For the first one take: $u=t^2$ and $dv=t\cdot e^{4t^2} dt$

For the second one take $u=t^2$ and $dv=e^{4t}$

**sbhatnagar** · December 6th 2011, 01:17 AM

$\int t^3(e^{4t^2}) \mathrm{d}t$

Substitute $x=t^2$ .

$\int t^3(e^{4t^2})\ \mathrm{d}t=\int \frac{x \cdot e^{4x}}{2}\ \mathrm{d}x$

$={1 \over 2}\left[ \frac{e^{4x}x}{4}-\frac{1}{4}\int e^{4x} dx\right]+C={1 \over 2}\left[ \frac{e^{4x}x}{4}-\frac{1}{16}e^{4x}\right]+C$

**paulbk108** · December 6th 2011, 01:31 AM

One of you folks mind completing one of them?

Once I get rolling I will be fine...

Regards.

**paulbk108** · December 6th 2011, 01:46 AM

sbhatnagar,

Can you please explain where the t^3 went? You substituted x = t^2...?

Regards.

**sbhatnagar** · December 6th 2011, 02:06 AM

I substituted $x=t^2$

$\frac{dx}{dt}=2t \iff dt=\frac{dx}{2t}$

$\int t^3{e^{4t^2}} dt=\int \frac{t^3(e^{4x})}{2t} dx=\int\frac{t^2(e^{4x})}{2} dx=\int \frac{x \cdot e^{4x}}{2} dx$

**Ackbeet** · December 8th 2011, 04:53 PM

Originally Posted by paulbk108

One of you folks mind completing one of them?

Once I get rolling I will be fine...

Regards.

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