MHB Solving Integration Help: Heartbeat During Workout

[rachelwind]

So here's the question:
If h(t) denotes the number of times a person’s heart beats in t minutes, then the pulse rate p beats per minute is given by p = dh/dt

During a workout an athlete has a pulse rate p(t) beats per minute where
p(t) = 58 + 10.62t^2(e^-0.25t)

t minutes after the start of the workout.

How many times does the athlete’s heart beat during the first 20 minutes of the workout? Give your answer to the nearest complete heartbeat.Ive been told i need two applications of integration by parts. But i really don't know how to do this. I've managed to do this so far:
Integral (58 dt) + 10.62 Integral ( t^2 e^(-t/4) dt )
58t + 10.62 Integral ( t^2 e^(-t/4) dt )

Let u = t^2. dv = e^(-t/4) dt.
du = 2t dt. v = (-4)e^(-t/4)

Getting the answer:
585 + (10.62)(-4)t^2 e^(-t/4) + (10.62)(8) Integral ( t e^(-t/4) dt )

Let u = t. dv = e^(-t/4) dt
du = dt. v = (-4)e^(-t/4)

Then:
585 + (10.62)(-4)t^2 e^(-t/4) + (10.62)(8)(-4)t e^(-t/4) - (10.62)(8)(-4) (-4)e^(-t/4) + CIs this right so far? Or am i completely off track. Thanks

 

[MarkFL]

Hello rachelwind, and welcome to MHB!

Let me first say that I truly appreciate the way you have shown your work and thoughts. This is very helpful to the helpers here in seeing just where you may need guidance. (Yes)

Now, I would agree (although using a slightly different notation), that if we let $H(t)$ be the total number of heartbeats after $t$ minutes, then we need to compute:

$$H(t)=\int_0^t p(x)\,dx$$

$$H(t)=\int_0^t\left(58+10.62x^2e^{-0.25x} \right)\,dx$$

$$H(t)=58\int_0^t\,dx+10.62\int_0^t x^2e^{-\frac{x}{4}}\,dx$$

Getting the simple first integral out of the way...

$$H(t)=58(t-0)+10.62\int_0^t x^2e^{-\frac{x}{4}}\,dx$$

$$H(t)=58t+10.62\int_0^t x^2e^{-\frac{x}{4}}\,dx$$

Now, I am curious how this first term becomes 585? Did you mean to type $58t$ since the "5" and the "t" keys are so close on the keyboard?

Looking at how you set up the integration by parts, you have proceeded correctly by choosing:

$$u=x^2\,\therefore\,du=2x\,dx$$

$$dv=e^{-\frac{x}{4}}\,dx\,\therefore\,v=-4e^{-\frac{x}{4}}$$

And so we have:

$$H(t)=58t+10.62\left(-4\left[x^2e^{-\frac{x}{4}} \right]_0^t+8\int_0^t xe^{-\frac{x}{4}}\,dx \right)$$

$$H(t)=58t+10.62\left(-4t^2e^{-\frac{t}{4}}+8\int_0^t xe^{-\frac{x}{4}}\,dx \right)$$

Now, you have chosen well in your second application of integration by parts in using:

$$u=x\,\therefore\,du=dx$$

$$dv=e^{-\frac{x}{4}}\,dx\,\therefore\,v=-4e^{-\frac{x}{4}}$$

And so we have:

$$H(t)=58t+10.62\left(-4t^2e^{-\frac{t}{4}}+8\left(-4\left[xe^{-\frac{x}{4}} \right]_0^t+4\int_0^t e^{-\frac{x}{4}}\,dx \right) \right)$$

$$H(t)=58t+10.62\left(-4t^2e^{-\frac{t}{4}}+8\left(-4te^{-\frac{t}{4}}-16\left[e^{-\frac{x}{4}} \right]_0^t \right) \right)$$

$$H(t)=58t+10.62\left(-4t^2e^{-\frac{t}{4}}+8\left(-4te^{-\frac{t}{4}}-16\left(e^{-\frac{t}{4}}-1 \right) \right) \right)$$

While I used the boundaries of the problem in the limits of integration, we have the same result (differing only by a constant), you would just need to use $H(0)=0$ to find $C$. Your working of the problem is flawless from what I see.

I would go on to clean my result up a bit to write:

$$H(t)=58t-42.48\left(t^2e^{-\frac{t}{4}}+8\left(te^{-\frac{t}{4}}+4\left(e^{-\frac{t}{4}}-1 \right) \right) \right)$$

Now it would just be a matter of evaluating $H(20)$ and rounding to the nearest integer. (Smile)

 

[Petrus]

Hello,

Nice done Mark!

Let me first say that I truly appreciate the way you have shown your work and thoughts. This is very helpful to the helpers here in seeing just where you may need guidance. (Yes)

I agree with you but look at your post! That some really well post and guide and especially effort!:) Well used latex cause I always screw somehow when It's a lot to write in latex :)

Is this right so far? Or am i completely off track. Thanks

Hello rachelwind,
Mark has answer your question but I would like to show you how I would solve, it's same as Mark but I do in a difrent way that makes it a lot easy for me cause It's big risk I confused myself when I integrate when you are suposed to integrate two function.

Regards,

 

[rachelwind]

Thankyou both. It has become much clearer to me now. (Handshake)