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Evaluate using the Fundamental Theorem of Calculus: $$\int_{-1}^1 \sqrt{1-x} \, dx$$

I'm not sure what to do with the $\sqrt{1-x}$. Most of the problems I'm doing for this assignment use substitution but I just end up getting $u = 1-x$ and of course $-du = dx$. Where do I go from here?

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What you've done seems helpful. Do you know the indefinite integral $-\int \sqrt{u}\,du = -\int u^{1/2}\,du$? Then you're almost done. –  Dylan Moreland Jul 25 '12 at 4:06
    
oh you know what, I didn't even realize that -du was -1du.. –  user69 Jul 25 '12 at 4:09

4 Answers 4

up vote 2 down vote accepted

Put $u=1-x\implies du=-dx$. Substituing it in integral and changing limits according to the relation $u=1-x$ converts your integral to $$-\int_2^0 \sqrt u du=\left.\frac{-2u^{3/2}}{3}\right|_2^0=4\frac{\sqrt 2}3$$

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Hint: $\sqrt{u}=u^{1/2}$. Can you go from there?

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Sorry! You must have answered while I was typing. –  Chris Leary Jul 25 '12 at 4:12

Can you integrate $\sqrt{u}$? As $x$ varies from $-1$ to $1$, what happens to $u=1-x$? You now have a small adjustment to make in the integral (what happens when you interchange the limits on the integral?), but it should be smooth sailing from here.

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Recall what the fundamental theorem of calculus says: If $f(x) = g'(x)$ on an interval $[a,b]$, then

$$\int_a^b f(x) = g(b) - g(a).$$

Now notice that $$\sqrt{1-x} = \frac{d}{dx} \Big(\frac{-2(1-x)^{3/2}}{3}\Big)$$

and applying the fundamental theorem with $f(x) = \sqrt{1-x}$, $g(x) = \Big(\frac{-2(1-x)^{3/2}}{3}\Big)$, $a= -1$ and $b = 1$ we get that

$$\begin{eqnarray*} \int_{-1}^1 \sqrt{1-x} &=& g(1) - g(-1) \\ &=& 0 - \frac{-2}{3}(1 - (-1))^{3/2} \\ &=& \frac{4\sqrt{2}}{3}. \end{eqnarray*}$$

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