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can someone help me solve this problem?

It is estimated that $t$-weeks into a semester, the average amount of sleep a college math student gets per day $S(t)$ decreases at a rate of $$\frac{-6t}{e^{t^2}}$$ hours per day. When the semester begins, math students sleep an average of 8 hours per day. What is $S(t)$, 8 week(s) into the semester?

Sorry, the equation is a bit weird, but please ignore that bracket.

I got until the point where 8 + the integral of it = unknown, and used substitution where I made $$u=e^{t^2}$$ and solved it to get: $$8+\frac{3}{u}$$ with 8 and 0 for coefficients. I tried solving it but the answer was not right. Can someone find my mistake? Thank you.

Toby Mak
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Hami the Penguin
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  • Can you please include all your work? Other people might give you an alternative solution, but it is hard to see where your mistake is from your work so far. – Toby Mak Aug 24 '19 at 04:35

1 Answers1

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I assume these are the steps you have followed.

Substitute $u = e^{t^2}$, $du = 2t e^{t^2} dt$ to get:

$$\int \frac{-6t}{e^{t^2}} \mathrm d t$$ $$=\int\frac{-6t}{u} \frac{du}{2t e^{t^2}} $$ $$= \int \frac{-3}{e^{t^2}} \frac{du}{u}$$ $$= -3 \int \frac{1}{u} \frac{du}{u}$$ $$= -3 \left(\frac{u^{-1}}{-1} \right) + C$$

Now the first problem: you have to substitute $u$ back in,

$$S(t) = 3e^{-t^2} + C$$

and the second problem is: $C$ is not $8$. To see why, substitute $t = 0$ into that last equation to find what $C$ is.

You should be able to continue.

Toby Mak
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