definite integrals

• Apr 21st 2013, 03:40 PM
sonalig
definite integrals
Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals http://calculus.sfsu.edu/latexrender...pe=png&size=15

Can somebody please help me with this one, I am not understanding how to approach it. I know it forms a fourth of a circle because of the bounds but that is all I am understanding.
• Apr 21st 2013, 03:47 PM
Plato
Re: definite integrals
Quote:

Originally Posted by sonalig
Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals http://calculus.sfsu.edu/latexrender...pe=png&size=15.

That is simply the area enclosed by the unit circle in the first quadrant.
How much is that?

• Apr 21st 2013, 06:36 PM
x3bnm
Re: definite integrals
Let's think of $x$ and $1$ in the given question as two sides of a triangle like in this image below:

Attachment 28076

$\cos{(\theta)} = \frac{x}{1} = x$

$\sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}$

$\text{And } dx = -\sin{(\theta)}\,\,d\theta$

Now for lower limit $x = 0$:
$x = \cos{(\theta)} = 0 \therefore \theta = \frac{\pi}{2}$

And for upper limit $x = 1$:

$x = \cos{(\theta)} = 1 \therefore\,\, \theta = 0$

Back to the main question which can be written as:

\begin{align*}\int_0^1 \sqrt{1-x^2}\,\,dx =& -\int_{\frac{\pi}{2}}^0 \sin^2{(\theta)}\,\,d\theta \\=& \int_{\frac{\pi}{2}}^0 \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2}\Bigg]_{\frac{\pi}{2}}^0 \\=& 0 - 0 - \frac{\sin{(\pi)}}{4} + \frac{\pi}{4} \\=& \frac{\pi}{4}....\text{[for 1st quadrant]}\end{align*}
• Apr 21st 2013, 06:53 PM
Plato
Re: definite integrals
Quote:

Originally Posted by x3bnm
Let's think of $x$ and $1$ in the given question as two sides of a triangle like in this image below:
Attachment 28076
$\cos{(\theta)} = \frac{x}{1} = x$
$\sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}$
$\text{And } dx = -\sin{(\theta)}\,\,d\theta$
Now for lower limit $x = 0$:
$x = \cos{(\theta)} = 0 \therefore \theta = \frac{\pi}{2}$
And for upper limit $x = 1$:
$x = \cos{(\theta)} = 1 \therefore\,\, \theta = 0$
Back to the main question which can be written as:
\begin{align*}\int_0^1 \sqrt{1-x^2}\,\,dx =& -\int_{\frac{\pi}{2}}^0 \sin^2{(\theta)}\,\,d\theta \\=& \int_{\frac{\pi}{2}}^0 \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2}\Bigg]_{\frac{\pi}{2}}^0 \\=& 0 - 0 - \frac{\sin{(\pi)}}{4} + \frac{\pi}{4} \\=& \frac{\pi}{4}....\text{[for 1st quadrant]}\end{align*}

@x3bnm
Why would anyone changed such a transparently simple question into such a complicated mess?
• Apr 21st 2013, 07:35 PM
x3bnm
Re: definite integrals
I guess I like complicated things (Happy) . Can you tell me how you would solve it if you don't mind telling me?
• Apr 22nd 2013, 01:54 AM
Prove It
Re: definite integrals
Quote:

Originally Posted by sonalig
Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals http://calculus.sfsu.edu/latexrender...pe=png&size=15

Can somebody please help me with this one, I am not understanding how to approach it. I know it forms a fourth of a circle because of the bounds but that is all I am understanding.

Surely you know that the area of a circle is calculated by $\displaystyle A = \pi \, r^2$. Here you have a quarter of the circle of radius 1, so what is the area?
• Apr 22nd 2013, 03:51 AM
agentmulder
Re: definite integrals
IMHO x3bnm has shown very nicely how to work integrals using trigonometric substitutions. The OP asked for sums and differences of areas which x3bnm provided using simple geometric formulas of right triangles.

:)
• Apr 22nd 2013, 05:00 AM
Prove It
Re: definite integrals
Quote:

Originally Posted by agentmulder
IMHO x3bnm has shown very nicely how to work integrals using trigonometric substitutions. The OP asked for sums and differences of areas which x3bnm provided using simple geometric formulas of right triangles.

:)

Surely an integral with trigonometric substitutions is much more difficult than $\displaystyle \frac{\pi \, r^2}{4}$. It's 100% correct and beautifully executed (and LaTeXed), but the question asked to use simple geometric formulas. Surely there are few more simple than the formula for the area of a circle...
• Apr 22nd 2013, 01:04 PM
x3bnm
Re: definite integrals
We know that the function of the given question is:

$y = \sqrt{1-x^2}........\text{(1)}$

By polar coordinate we know that $x = r\cos(\theta)$ and $y = r\sin(\theta)$
If we plugin these values into equation (1) we get:

\begin{align*}r\sin{(\theta)} =& \sqrt{1-r^{2}\cos^2{(\theta)}} \\ r^2\sin^2{(\theta)} =& 1 - r^{2}\cos^2{(\theta)}\\ r^2(\sin^2{(\theta)} + \cos^2{(\theta)}) =& 1\\ \therefore\,\, r =& 1\end{align*}

For upper limit $x = 1$
$x = r\cos{(\theta)} = \cos{(\theta)} = 1 \therefore\,\, \theta = 0$

For lower limit $x = 0$

$x = r\cos{(\theta)} = \cos{(\theta)} = 0 \therefore\,\, \theta = \frac{\pi}{2}$

So the area is one fourth of the full quadrant.

$Area = \pi r^2 - \frac{3 \pi r^2}{4} = \frac{\pi}{4}.........\text{[because } r = 1 \text{]}$