1

I'm reading Thomas calculus on page 657. There is an integral of a function

$$\frac{45}{2000-5t}$$

And I found the integration of

$$\frac{45}{2000-5t}$$

is totally different from

$$\frac{9}{400-t}$$

, which I think are

$$-9\cdot\ln(|2000-5t|)$$

and

$$-9\cdot\ln(|400-t|)$$

respectively. I think

$$\frac{45}{2000-5t}$$

should be equal to

$$\frac{9}{400-t}$$

so does their integral. Did I do anything wrong or make a stupid mistake, or there is some special knowledge of this kind of integration?

Any help will be appreciated. Thank you!

M. Chen
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2 Answers2

8

The two functions differ by a constant:

$$-9 \ln|2000-5t| = -9\ln(5\cdot |400-t|) = -9(\ln(5) + \ln|400-t|) = \ln|400-t| + C$$

for some constant $C$ (that constant is actually equal to $-9\ln5$).

Remember, when calculating the indefinite integral, the teachers aren't just being annoying when they say don't forget the constant.

In fact, when you are looking at $\int f(x)dx$, you are in fact looking at all the functions whose derivative is $f$. We represent this set of functions using a free parameter $C$ which is a constant. So, for example, $$\int 2x dx = x^2 + C$$ because the set $$\{x\mapsto x^2 + C| C\in\mathbb R\}$$ is the set of all functions whose derivative is equal to $2x$.

In your case, the sets

$$\{x\mapsto \ln|2000-5t| + C| C\in \mathbb R\}$$

and

$$\{x\mapsto \ln|400-t|+C| C\in\mathbb R\}$$

are the exact same set, so both answers are equally valid.

5xum
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3

Result of integration is not unique function, it is unique up to an additive constant. Since logarithm of a product is sum of logarithms $$log(a\cdot b)=log (a) + log (b)$$ you get that $$log(2000-5t)=log (400-t)+log(5)$$ and $log (5)$ is a constant.

TStancek
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