Calculate the integral of ....

Question

$\int_{7}^{10}\sqrt{-40+14x-x^2}dx$

I started off by doing $\int_{7}^{10}\sqrt{(x+7)^2-89}dx$ but I don't know whether that's correct and how I should proceed.

Edit: Ok so it should be $\int_{7}^{10}\sqrt{9-u^2}dx$ with u = x+7. How do I factor out the 9 to get something in the form of $\int_{7}^{10}\sqrt{1-u^2}dx$?

Perhaps factoring inside the radical will help: $-40+14x-x^2 = -(x-4)(x-10)$. — Eff, Oct 16 '14 at 21:07
You need $9-(x-7)^2$ under the square root - expand and check. — Mark Bennet, Oct 16 '14 at 21:07
And once you've fixed the arithmetic error, you will want to think about a trigonometric substitution for $x+7$. — rogerl, Oct 16 '14 at 21:08
http://math.stackexchange.com/questions/390080/definite-integral-of-square-root-of-polynomial — hickslebummbumm, Oct 16 '14 at 21:17

score 2 · Answer 1 · answered Oct 16 '14 at 21:10

2

Methodology answer:

In principle, completing the square is the right start. But you didn't complete the square correctly. Notice, for instance, that it started off with the term $-x^2$, and in your rewrite you have $x^2$.

But once you fix that, you might do the substitution $u = x - 7$. You will then have an integral of the form

$$\int_a^b \sqrt{9 - x^2} dx.$$

In these problems, you need to somehow find a different expression for the radical term. You should probably use trig substitution (although if you really wanted, you could use other techniques too).

answered Oct 16 '14 at 21:10

davidlowryduda

91,687

Ok so if I got $\int_{7}^{10}\sqrt{9-u^2}dx$, how can I factor the 9 out of the radical so I get the form 1-x^2? Nice site by the way – CheekyKontBrah Oct 16 '14 at 21:39
@Ken: You could just factor it out. This would lead you to $3\sqrt{1 - (u/3)^2}$, and you could go from there. – davidlowryduda Oct 16 '14 at 22:06

Calculate the integral of ....

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