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I am currently reading Tom Apostol's Calculus Vol.1 and I am somehow stuck on this problem: Having defined $$\pi = 2 \int_{-1}^{1}{\sqrt{1-x^2}dx}~~,$$ use the properties of the integral to compute the following in terms of $~\pi~$: $$\int_{-3}^{3}{\sqrt{9-x^2}dx}$$ The properties that have been shown so far in this book are: $$\int_a^b{\sum_{k=1}^n{c_kf_k(x)dx} = \sum_{k=1}^nc_k\int_a^bf_k(x)dx}$$ $$\int_a^b{f(x)dx}+\int_b^c{f(x)dx} = \int_a^c{f(x)dx}$$ $$\int_a^b{f(x)dx}=\int_{a+c}^{b+c}{f(x-c)dx}$$ $$\int_a^b{g(x)dx} \leq \int_a^b f(x)dx\;\text{If}\; g(x) \leq f(x)$$ Along with proofs of these theorems.

Edit from OP: I have forgotten a property like a dummy. The missing one is: $$\int_a^b{f(x)}dx={\frac{1}{k}\int_{ka}^{kb} f(\frac{x}{k})dx}$$ Sorry for the mistake.

Now since the integrand here is represented by a semi-circle of radius $~3~$, it is fairly obvious that the anwser is $~\frac{9\pi}{2}~$ .

Yet I can't seem to be able to use the aforementioned properties to find it in the way that author wants me to.

I assumed that I could somehow derive the following formula : $$9\int_{-1}^1\sqrt{1-x^2}dx = \int_{-3}^{3} \sqrt{9-x^2}dx$$ And it would follow that the anwser was $\frac{9\pi}{2}$ ,but I keep hitting dead ends.

Could somebody please explain to me the method (involving only the aforementioned properties) that should be used to solve this?

I've been sitting on this for a while and I'm wondering if this is simply a case of not being able to find the right numbers or if I am simply not on the right track.

P.S.: This is my first post here/first time using Mathjax, so please excuse/point out any mistakes I might have made in the formulas.

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    $\begingroup$ Frankly I'm going to be honest with you and say I'm not really sure either. The obvious idea is to use linear u-substitution:$$\int_{ma}^{mb}mf(mx)~\mathrm dx=\int_a^bf(x)~\mathrm dx$$which is straightforward to prove from Riemann sums. Looking at the given integral properties however, I'm curious as to how the last one might play into this. My intuition says squeeze theorem, but first we need to convert the bounds from $[-3,3]$ to $[-1,1]$, which can only be done by the second property; none of the other properties change the size of the interval we are working with. $\endgroup$
    – Simply Beautiful Art
    Jul 17, 2019 at 2:35

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With those properties is not possible to prove this, in fact you could prove first this one : $$\int_{ca}^{cb} f(x)dx =c \int_a^b f(cx)dx \quad , \ c\geq 0$$ After that, note that $$\int_{-3}^3 \sqrt{9-x^2}\ dx = \int_{-3}^3 \sqrt{9\left( 1-\frac{x^2}{9} \right)} \ dx =3 \int_{-3}^3 \sqrt{1-\left( \frac{x}{3} \right)^2 }\ dx$$ and apply the previous "property" for $c=3$.

Finally, you would end up with the latter being equal to $$9 \int_{-1}^1 \sqrt{1-x^2} \ dx = \frac{9\pi}{2}$$

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    $\begingroup$ This disagrees with the answer given (by the area of a circle) and the formula is wrong. $\endgroup$
    – Simply Beautiful Art
    Jul 17, 2019 at 2:39
  • $\begingroup$ Already corrected $\endgroup$
    – azif00
    Jul 17, 2019 at 2:40
  • $\begingroup$ It's not correct still... $\endgroup$
    – Simply Beautiful Art
    Jul 17, 2019 at 2:40
  • $\begingroup$ ... Why not? ..... $\endgroup$
    – azif00
    Jul 17, 2019 at 2:42
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    $\begingroup$ Ah my bad. For some reason it didn't refresh your answer to the properly updated version. $\endgroup$
    – Simply Beautiful Art
    Jul 17, 2019 at 2:47

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