## Monday, January 15, 2007

### What Can Integrals Do For You? Topic: Calculus.

Problem: Evaluate $\displaystyle \int_{-1}^0 \sqrt{\frac{1+x}{1-x}} dx$.

Solution: Well, it looks not cool right now, so lets multiply through by $\displaystyle \sqrt{\frac{1+x}{1+x}}$. We get $\displaystyle \int_{-1}^0 \frac{1+x}{\sqrt{1-x^2}} dx$.

That's nice; split up the integral into two parts, from which we obtain

$\displaystyle \int_{-1}^0 \frac{1}{\sqrt{1-x^2}} dx + \int_{-1}^0 \frac{x}{\sqrt{1-x^2}}dx = \left[\arcsin{x}-\sqrt{1-x^2}\right]_{-1}^0 = \frac{\pi}{2}-1$.

QED.

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Comment: Not the toughest of integrals, but if you tried various substitutions like I did first, you probably found yourself with worse ones. Funny how a simple thing like this solves it.

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Practice Problem: Let $f(x)$ be a differentiable and increasing function such that $f(0) = 0$. Prove that $\displaystyle \int_0^1 f(x) f^{\prime}(x)dx \ge \frac{1}{2}\left(\int_0^1 f(x) dx \right)^2$.

1. omg I get this one

Practice problem: integration by parts. so u=f(x) dv=derivative of f(x)
and u solve.... umm....

well, wut about if you take the f(x) out of the integral, then the integral of the derivative would be f(x) so the first whole part is (f(x))^2, and the second part, it's the area of f(x) between 0 and 1, so .. i know starts from the origin, and the area squared... and take half of that, which is probably (f(x)) ^2 times 1/2, so the second whole part is only half of the first blob.

ok, i know there's gotta be something wrong in that shpeel, but the practice problem has nothing to do with your real problem. i get the arcsin part...

2. Lol. You can't take f(x) out of the integral. But a good substitution for the left side would be u = f(x), du = f'(x)dx.

And for the right side, think about what you know if f(x) is increasing and try to match it with what you got from the left side.

3. substitution's a good idea haha so the integral becomes 1/2 (int f(x))^2 from 0 to 1 (which is not the same as 1/2 (int f(x) from 0 to 1)^2 of course)

let int f(x) = F(x), then the LHS is 1/2 F(1)^2 - F(0)^2 whereas the RHS is 1/2 ( F(1) - F(0) )^2, and since f(0) = 0 and f(x) is increasing then -2 F(1) F(0) is nonnegative, so the result is obvious

4. how can you conclude that f(x) is increasing just by the point f(0)=0?

5. "Let f(x) be a differentiable and increasing function..."